1 Real Analysis
Real analysis is the rigorous study of functions, sequences, and series defined on the real numbers, built on the foundational property of completeness of \(\mathbb{R}\). It begins with limits of sequences and functions, convergence tests, and continuity, and then develops differentiation and integration from an \(ε–δ\) perspective. A central theme is understanding how infinite processes (limits, series, integrals) behave, with distinctions between pointwise and uniform convergence. Real analysis also extends classical calculus through topics such as metric spaces, compactness, connectedness, and measure theory, culminating in the Lebesgue integral. These tools provide the framework for modern analysis, probability theory, and the rigorous treatment of differential equations and functional spaces.
1.1 Preliminary Material
1.1.1 The Real Numbers
The reals are a field and thus satisfy the field axioms, that is there exists operations addition and subtraction that are commutative, associative and have identities. The real numbers also have a natural order relation that is compatible with their algebraic structure. We visualize the real numbers as the real line, with smaller numbers to the left and larger numbers to the right.
There is a strict linear order \(<\) on \(\mathbb{R}\) such that for all \(x,y,z \in \mathbb{R}\): 1. either \(x<y\), \(x=y\) or \(x>y\); 2. if \(x<y\) then \(x+z<y+z\); 3. if \(x<y\) and \(z>0\), then \(xz<yz\).
The real numbers are also Dedekind Complete, that is they satisfy the least-upper-bound property. Completeness implies that there are no”gaps” or “missing points” in the real number line.
The real numbers \(\mathbb{R}\) are Dedekind complete since every nonempty subset of \(\mathbb{R}\) that is bounded above has a least upper bound (supremum) in \(\mathbb{R}\).
1.1.2 Real Sequences
We typically consider the extended real numbers \(\overline{\mathbb{R}}=\mathbb{R}\cup \{ \pm \infty \}\) such that every subset \(A\subset\overline{\mathbb{R}}\) has a supremum and infimum. From completeness, it follows that every sequence \(\{ x_{n} \}_{n \in \mathbb{N}}\subset\overline{\mathbb{R}}\) has a limit superior and a limit inferior given by \[ \limsup_{n\in \mathbb{N}}{x_{n}}=\inf_{k \geq 1}{\left( \sup_{n \geq k}{x_{n}} \right) }\quad \&\quad\liminf_{n \in \mathbb{N}}{x_{n}}=\sup_{k \geq 1}{\left( \inf_{n \geq k}{x_{n}} \right) }. \] :::{.definition data-title=“Convergence of Real Sequences”}
A sequence \(\{ x_{n} \}_{n \in \mathbb{N}}\subset \overline{\mathbb{R}}\) converges to limit \(x \in\overline{\mathbb{R}}\), denoted by \(\lim_{ n \to \infty }x_{n}=x\) or \(x_{n}\to x\), as \(n\to \infty\) if \[ \forall \epsilon >0, \exists N_{\epsilon}\in \mathbb{N} : \forall n > N_{\epsilon}, \lvert x_{n}-x \rvert <\epsilon. \]
:::
This is a special case of a more general type of functional, or indeed topological convergence, known as \(\epsilon\)-\(\delta\) convergence discussed in later sections.
The sequence \(\{ x_{n} \}\) converges iff \(\limsup_{}{x_{n}=\liminf_{}{x_{n}}}\) and in this case the limit is given by \[ \lim_{ n \to \infty } x_{n}=\limsup_{}{x_{n}}=\liminf_{}{x_{n}}. \] :::{.definition data-title=“Cauchy Sequence”}
A real sequence \(\{ x_{n} \}\) is Cauchy if \[\forall \epsilon >0, \exists N=N_{\epsilon}~s.t.~|x_{n}-x_{m}|>\epsilon ,\forall n,m>N,\]or in other words \(x_{n}-x_{m}\to 0\) as \(n,m \to \infty\) independently.
:::
A real sequence \(\{ x_{n} \}\) is convergent in the reals if and only if it is Cauchy.
1.1.3 Topology
In this section we highlight some important results from our study of Topology.
A topology \(\tau\) on general set \(X\) is a collection of subsets of \(X\) satisfying the following propoerties: 1. \(\emptyset \in \tau\) and \(X \in \tau\); 2. \(\{ U_{i} \}_{i \in \mathcal{I}}\subset X\implies \bigcup_{i \in\mathcal{I}}U_{i}\subset X\) (closed under arbitrary unions); and 3. \(\{ U_{i} \}_{i=1}^N\subset X \implies \bigcap_{i=1}^N U_{i}\subset X\) (closed under finite intersections).
All elements \(E\in \tau\) are called open sets.
A couple consisting of a set and an assigned topology \((X,\tau)\) is known as a topological space. Note that a topological space need not have a formal definition of distance or size and so we are limited to considering subsets called neighborhoods.
For a point \(x \in X\) a neighborhood is a subset \(V\subset X\) that contains an open set \(U\) such that \(x \in U\).
1.1.4 Metric Spaces
In real analysis we need not be so general and typically work with specific types of topological spaces known as metric spaces upon which we have defined some function describing distance between points.
A metric \(d:X\times X \to \mathbb{R}^+\), a function that satisfies the following for all \(x,y,z\in X\): 1. \(d(x,x)=0\)
2. \(x \neq y\implies d(x,y)\neq 0\) 3. \(d(x,y)=d(y,x)\) 4. \(d(x,z)\leq d(x,y)+d(y,z)\)
A metric space is an ordered pair \((X,d)\) consisting of a set \(X\) and a metric \(d\).
The standard metric for the reals \(\mathbb{R}^n\) is the Euclidean metric which is given by \[ d(x,y)=\left( \sum_{n}^{}(x_{n}+y_{n})^2 \right)^{1/2}. \] The open ball at point \(c \in X\) of diameter \(\epsilon\) is defined as \[ B_{\epsilon}(c):=\{ x \in X:d(x,c)<\epsilon \}. \] ### Real Functions
Let \(X,Y\subset \overline{\mathbb{R}}\). A real function is a mapping \(f:X \to Y\). Recall that a function is injective (one-to-one) when \[ x_{1}\neq x_{2} \implies f(x_{1})\neq f(x_{2}), \] and surjective (onto) if \[ \forall y \in Y, \exists x \in X:f(x)=y. \] A function which is both injective and surjective is said to be bijective. A bijection is invertible, that is there exists function \(f^{-1}\) called the inverse such that \[ \forall x \in X, f\circ f^{-1}(x)=f^{-1}(f(x))=x. \] A function is increasing if \(x \leq y \implies f(x) \leq f(y)\) and decreasing if \(x\leq y\implies f(x)\geq f(y)\). These become strict should we drop the equality. A function that is either increasing or decreasing is called monotone.
A function is continuous when a small variation of the argument induces a small variation of the value of the function.
A function \(f:X \to Y\) is continuous at point \(c\) if \[ \forall\epsilon>0,\exists \delta>0:\forall x \in B_{\delta}(c), f(x)\in B_{\epsilon}(f(c)). \]
This is equivalent to writing that \(\lim_{ x \to c }f(x)=f(c)\). Simplifying to considering the real functions with the standard Euclidean metric \(d(x,y) = \lvert x-y \rvert\) we obtain the familiar definition given below.
A real function \(f:X \to Y\) is continuous at point \(c\) if \[ \forall \epsilon>0, \exists \delta>0 : (\forall x \in S: \lvert x-c \rvert <\delta), \lvert f(x)-f(c) \rvert <\epsilon. \]
Intuitively, no matter how small we make \(\epsilon\) we can find a neighborhood such that for all points within the region (called interior points), the distance between points in their image is less than \(\epsilon\).
Assuming that the functions \(f,g\) are both continuous at \(c\) the following combinations are continuous: 1. \((f+g)(x)=f(x)+g(x)\); 2. \((f-g)(x)=f(x)-g(x)\); 3. \((f\cdot g)(x)=f(x)\cdot g(x)\); 4. \(\left( \frac{f}{g} \right)(x)=\frac{f(x)}{g(x)}\) if \(g(c)\neq 0\); 5. \(\lambda f(x)\) for constant \(\lambda\); and 6. \((f \circ g)=f(g(x))\) if \(f\) is continuous at \(g(c)\).
We say that the function \(f\) is right-continuous and left-continuous respectively when \[ \lim_{ x \to c^+ } f(x)=f(c)\quad \&\quad \lim_{ x \to c^-} f(x)=f(c), \] that is, the limit is obtained when \(x\) approaches \(c\) from the right or the left respectively.
Another important property of functions is upper semicontinuity (usc) and lower semicontinuity (lsc).
A function \(f:X\to \mathbb{R}\) is upper semicontinuous at \(c\) if \[ \forall\epsilon>0,\exists \delta>0:\forall x \in B_{\delta}(c),f(x)-f(c)<\epsilon. \] A function \(f:X\to \mathbb{R}\) is lower semicontinuous at if \[ \forall\epsilon>0,\exists \delta>0:\forall x \in B_{\delta}(c),-\epsilon<f(x)-f(c). \]
Notice that these definitions are two halves of the condition of continuity, namely that we have taken \[ \lvert f(x)-f(c) \rvert<\epsilon \implies -\epsilon<f(x)-f(c)<\epsilon, \] where the left inequality goes to lsc and the right inequality goes to usc. Clearly therefore a function is continuous if and only if it is both upper and lower semicontinuous.
We list some equivalent definitions of usc at point \(c\) below: 1. \(\{ x:f(x)<c \}\) is open 2. \(f(c)\geq \limsup_{n \to \infty}{f(x_{n})}\) for all \(x_{n}\to c\)
We list some equivalent definitions of lsc at point \(c\) below: 1. \(\{ x:f(x)>c \}\) is open 2. \(f(c)\leq \liminf_{n \to \infty}{f(x_{n})}\) for all \(x_{n}\to c\)
These alternative definitions are helpful in our study of measure theory, particularly for showing that lsc and usc functions are measurable with respect to the Borel \(\sigma\)-algebra.
Given a metric space \((X,d)\) and a function \(f:X\to\overline {\mathbb{R}}\) the lower semicontinuous envelope of \(f\) is defined by \[ \begin{align}f_{*}(x):= & \lim_{\epsilon \to 0}\left(\inf\{ f(y):d(x,y)<\epsilon \}\right) \\ = & \lim_{ \epsilon \to 0 }(\inf_{}{f(B_{\epsilon}(x))}).\end{align} \]
1.1.5 Homeomorphisms
In topology, a homeomorphism is a bijective and continuous function between topological spaces that has a continuous inverse function. Speaking generally, a topological space is a geometric object and a homeomorphism results from a continuous deformation of the object into a new shape.
A function \(f:X\to Y\) between topological spaces is a homeomorphism if: 1. \(f\) is a bijection; 2. \(f\) is continuous; and 3. \(f^{-1}\) is continuous.
If such a function exists then spaces \(X\) and \(Y\) are said to be homeomorphic which is an equivalence relation on topological spaces. Thus, two homeomorphic spaces share the same topological properties e.g. compactness, connectedness, Hausdorff etc.
Let \(f:[a,b]\to \mathbb{R}\) be continuous on the closed interval \([a,b]\). If \(L\) is any real number between \(f(a)\) and \(f(b)\) then there exists some \(c\in[a,b]\) such that \[ f(c)=L. \]
Proof: Defining \(S:=\{x\in[a,b]:f(x)\leq L\}\) we note that \(a\in S\) and \(S\subseteq[a,b]\), therefore \(S\) is non-empty and bounded above. We define \(c:=\sup S\) and claim that \(f(c)=L\). \((\leq)\) Assuming that \(f(c)>L\), by the continuity of \(f\) at \(c\), there exists \(\delta >0\) such that \[
|x-c|<\delta \implies f(x)>L.
\] Thus for all \(x<c\) sufficiently close to \(c\), we have that \(f(x)>L\) contradicting the fact that \(c=\sup S\) (since there must be points of \(S\) arbitrarily close to \(c\) from below). \((\geq)\) Assuming that \(f(c)<L\), by the continuity of \(f\) at \(c\) there exists \(\delta > 0\) such that \[
|x-c|<\delta \implies f(x)<L.
\] Thus for all \(x>c\) sufficiently close to \(c\) we have that \(f(x)<L\) contradicting that \(c\) is an upper bound of \(S\).
Combining both inequalities we conclude that \(f(c)=L\). \(\square\)
Let \(a\in\mathbb{R}\). Suppose \(f,g,h\) are functions defined on some punctured neighborhood of \(a\) (i.e. for all \(x\) with \(0<|x-a|<\delta\) for some \(\delta>0\)). Assume that: 1. \(f(x)\leq g(x)\leq h(x)\) for all \(x\) sufficiently close to \(a\) with \(x\neq a\). 2. \(\lim_{x\to a}f(x) = \lim_{x\to a}h(x)=L\).
Then \(\lim_{x\to a}g(x)=L\).
Proof: Let \(\varepsilon > 0\). Since \(\lim_{x \to a} f(x) = L\) there exists \(\delta_1 > 0\) such that \[ 0 < |x - a| < \delta_1 \implies |f(x) - L| < \varepsilon, \] which implies \(L - \varepsilon < f(x)\). Similarly, since \(\lim_{x \to a} h(x) = L\) there exists \(\delta_2 > 0\) such that \[ 0 < |x - a| < \delta_2 \implies |h(x) - L| < \varepsilon, \] which implies \(h(x) < L + \varepsilon\). By assumption, there exists \(\delta_3 > 0\) such that \[ 0 < |x - a| < \delta_3 \implies f(x) \le g(x) \le h(x). \] Now let \(\delta := \min\{\delta_1, \delta_2, \delta_3\}\). Then for all \(x\) with \(0 < |x - a| < \delta\), \[ L - \varepsilon < f(x) \le g(x) \le h(x) < L + \varepsilon. \] Hence, \(|g(x) - L| < \varepsilon\). Since \(\varepsilon > 0\) was arbitrary, it follows that \(\lim_{x \to a} g(x) = L\). \(\square\)
1.1.6 Reimann Integration
A function \(f(x)\) is differentiable at point \(c\) if the limit of the difference quotient exists and is finite, that is \[ f'(c)=\lim_{ n \to \infty } \frac{{f(x)-f(c)}}{x-c}\in \mathbb{R}. \]
1.2 Measure Theory
In real analysis in order to handle complicated real functions we need to turn to Lebesgue integrals and Measure Theory. The primary aim of Measure Theory is to define the concept of measurability and ultimately assign a measure to measurable sets. That is, for general \(X\) we wish to construct a function \(\mu:2^{X}\to[0,\infty]\) that assigns each \(E\subset X\) a number \(\mu(E)\in[0,\infty]\), the measure of \(E\) where \(\mu\) satisfies:
- For disjoint \(E_{1}, E_{2}, \dots \in X\) we have \(\mu\left( \bigcup_{j=1}^\infty E_{j} \right)=\sum_{j=1}^\infty \mu(E_{j})\) (countably additive);
- For interval \([a,b]\) we have \(\mu([a,b])=b-a\); and
- For set \(E\subset \mathbb{R}\) and constant \(c\in \mathbb{R}\) we have \(\mu(E+c)=\mu(E)\) (translation invariance).
1.2.1 Motivations and Challenges
The Vitali Non-Measurability Theorem demonstrates that no such function exists. Indeed the closest mathematical generalization of mass is σ-additivity which gives rise to the Lebesgue measure, however not all sets have such a measure (for example, a Vitali set). The idea of the theorem can be extended to higher dimensions (see Banarch-Tarski Theorem) as well as similar results for other abstract sets although the form of the proof relies heavily on the nature of the set being considered.
We preemptively prove the following lemma proving that we obtain monotonicity from countable additivity of functions.
Given a set \(X\) and countably additive function \(\mu:2^X\to[0,\infty]\) then for all \(A,B\subseteq X\) \[ A \subseteq B\implies \mu(A)\leq \mu(B). \]
Proof: Since \(A \subseteq B\) we can write \(B =A\cup(B\setminus A)\implies \mu(B)=\mu(A)+\mu(B\setminus A)\geq \mu(A)\) since measures are non-negative. \(\square\)
There exists no function \(\mu :2^\mathbb{R}\to[0,+\infty]\) that is countably additive, translation invariant and satisfies \(\mu([a,b])=b-a\) for all \(a\leq b\).
Proof: Assume for the sake of contradiction that such a \(\mu\) exists. We define an equivalence relation \(\sim\) on \(\mathbb{R}\) \[ x\sim y\iff x-y \in \mathbb{Q}. \] The equivalence class of \(\sim\) to which \(x\) belongs is therefore given by \[ [x]_{\sim}:=\{ y \in \mathbb{R} : x\sim y \}. \] We make the claim that every such equivalence class contains an element in \([0,1]\). Assuming the claim is false then \(\exists x \in \mathbb{R}\) s.t. \([x]_{\sim}\cap[0,1]=\emptyset \implies \forall a \in [0,1],~x-a\not\in \mathbb{Q}\) i.e. \([x-1,x]\) contains no rational numbers which is impossible since the rationals are dense in the reals, giving the contradiction. Using the [[axiom-of-choice]], for each equivalence class we choose one element in \([0,1]\) belonging to that class and call the resulting set \(A\). Then we define \[ A_{q}=\{ A+q \}_{q \in \mathbb{Q}\cap [-1,1]}=\{ a+q:a\in A,q\in \mathbb{Q}\cap[-1,1] \}, \] hence each \(A_{q}\) contains the elements of \(A\) shifted by rational \(q \in [-1,1]\). We assume these are not disjoint i.e. \(\exists x \in A_{q}\cap A_{p}~s.t.~p \neq q\). This gives that \[ \begin{cases} x = a+p\quad\therefore\quad a=x-p \in A \\ x = a+q\quad \therefore\quad a = x-q \in A, \end{cases} \] \(\implies x-p\) and \(x-q\) are distinct elements of \(A\) from the same equivalence class giving the contradiction \(\rightarrow\leftarrow\). We therefore define the disjoint union \(B = \bigcup_{q \in\mathbb{Q}\cap [-1,1]}A_{q}\) and claim that \[ [0,1]\stackrel{(i)}{\subseteq} B \stackrel{(ii)}{\subseteq} [-1,2]. \] (i) Since \(\forall x \in \mathbb{R}\) we have \(x \in [a]\) for some \(a \in A\), in particular \(\forall x \in [0,1]\), \(x -a\in \mathbb{Q}\) for some \(a \in A\), i.e. \(x = a+q\) for some \(a \in A\) and \(q \in \mathbb{Q}\). Since \(x \in[0,1]\) and \(a \in A \subseteq [0,1]\), \(q = x-a\in[-1,1]\) and thus \(x \in B\). (ii) If \(b \in B\), then \(b = a+q\) for \(a \in A \subseteq [0,1]\), \(q \in [-1,1]\) and so \(b \in [-1,2]\). By monotonicity of countably additive functions \[ 1 = \mu([0,1])\leq \mu(B)\leq \mu([-1,2])=3. \] However, we also note that \[ \mu(B)=\sum_{q\in\mathbb{Q}\cap [-1,1]}\mu(A+q)=\sum_{q \in \mathbb{Q}\cap [-1,1]}\mu(A). \] Since \(\mu(B)\leq 3\) we must have that \(\mu(A)=0\), however this implies \(\mu(B)=0\) which contradicts our previous conclusion that \(\mu(B)\geq 1\). \(\rightarrow\leftarrow\) \(\square\)
To overcome this problem we turn to some standard approaches for mathematicians, either (1) weaken the criteria; or (2) restrict the domain of the function to the family of well-behaved sets. Both options are valid and we will see that (1) leads us to the concept of outer-measure, and (2) leads to the concept of measure spaces and measures.
Let us consider how we might restrict the domain of the function and what kind of family of subsets we should restrict to. The answer is a non-empty collection of subsets that is both closed under compliments and closed under countable unions (i.e. sets that preserve the properties we wish our measure to have) also known as a \(\sigma\)-algebra, a specific type of algebra.
1.2.2 Algebra
An algebra is a collection of sets that is non-empty, and closed under finite unions and compliments.
Let \(X\) be a non-empty set. An algebra of sets on \(X\) is a collection \(\mathcal{A}\subseteq 2^X\) of subsets of \(X\) that satisfies: 1. \(X \in \mathcal{A}\) 2. \(A,B\in \mathcal{A} \implies A\cup B\in \mathcal{A}\) (closed under finite unions); and 3. \(A\in \mathcal{A}\implies A^c:=X\setminus A\in \mathcal{A}\) (closed under compliments).
Some examples of algebra include: (1) the trivial sigma algebra \(\{ \emptyset,X \}\); (2) the power set \(2^X\); and (3) the set of all finite and cofinite subsets of \(X\) defined \(\mathcal{A}:=\{ E\subseteq X:\#E<\infty~\text{or}~\#X\setminus E < \infty \}\).
If \(\mathcal{A}\) is an algebra then: (1) \(\emptyset \in\mathcal{A}\), \(X\in \mathcal{A}\); and (2) \(E_{1}, \dots, E_{n}\in \mathcal{A}\implies \bigcap_{i=1}^nE_{i}\in\mathcal{A}\).
PROOF: (1) Since \(\mathcal{A}\) is non-empty we have set \(A\in\mathcal{A}\implies A^c\in\mathcal{A}\) \(\implies X:=A\cup A^c\in\mathcal{A}\) \(\implies \emptyset=X^c\in\mathcal{A}\). (2) \(E_{1}, \dots, E_{n}\in \mathcal{A}\) \(\implies E_{1}^c, \dots, E_{n}^c\in \mathcal{A}\) \(\implies \bigcup_{i=1}^nE_{i}^c\in\mathcal{A}\) \(\implies \bigcap_{i=1}^nE_{i}=\left(\bigcup_{i=1}^n E_{i}^c\right)^c\in \mathcal{A}\). \(\square\)
Note: In mathematical etymology we typically use the prefix σ to denote that the object is closed with respect to countable unions.
A σ-algebra \(\mathcal{F}\) on \(X\) is an algebra that is closed under countable unions, that is \[ \{ A_{n} \}_{n=1}^\infty \subseteq \mathcal{F} \implies \bigcup_{n} A_{n}\in \mathcal{F}_{n}. \]
We say that the pair \((X, \mathcal{F})\) consisting of a non-empty set \(X\) and a \(\sigma\)-algebra is a measurable space and that any set \(E\in \mathcal{F}\) is measurable.
A \(\sigma\)-algebra \(\mathcal{F}\) is closed under countable intersections.
PROOF: The result follows using a methodology identical to (2) in the previous result for algebra. \(\square\)
Any algebra that is closed under countable disjoint unions must be a \(\sigma\)-algebra.
PROOF: For countable (possibly non-disjoint) sets \(\{ A_{n} \}_{n=1}^\infty \subseteq \mathcal{A}\) we can define the sets \(B_{1}=A_{1}\) and \(B_{n}=A_{n}\setminus B_{n-1}\) such that \(\{ B_{n}\}_{n=1}^\infty\) are disjoint and \(\cup_{n} A_{n}=\cup_{n}B_{n}\in\mathcal{A}\). \(\square\)
Given any \(\mathcal{E}\subseteq 2^X\) there exists a smallest \(\sigma\)-algebra containing \(\mathcal{E}\) denoted \(\sigma(\mathcal{E})\) known as the \(\sigma\)-algebra generated by \(\mathcal{E}\). That is, for all \(\mathcal{F}\) containing \(\mathcal{E}\) then \(\sigma(\mathcal{E})\subseteq \mathcal{F}\).
PROOF: Naturally, the intersection of all \(\sigma\)-algebra that contain \(\mathcal{E}\) will infimum of all such \(\sigma\)-algebra and so it suffices to show that the intersection of \(\sigma\)-algebra is itself a \(\sigma\)-algebra. For set \(X\) consider the collection of \(\sigma\)-algebra \(\{ \mathcal{F}_{n} \}_{n=1}^\infty\) and denote the intersection \(\mathcal{F}:=\bigcap_{n=1}^\infty \mathcal{F}_{n}\). Clearly \(\emptyset \in \mathcal{F}_{n},~\forall n\) hence \(\emptyset \in \mathcal{F}\) is non-empty. Additionally, if \(A\in \mathcal{F}\) then \(A\in \mathcal{F}_{n},~\forall n\) hence \(A^c \in \mathcal{F}_{n},~\forall n\) and so \(A^c\in \mathcal{F}\). Finally, let \(\{ A_{m} \}_{m=1}^\infty \in \mathcal{F}\) therefore \(A_{m}\in \mathcal{F}_{n},\forall n.m\) hence \(\bigcup_{m}A_{m}\in \mathcal{F}_{n},~\forall n\) and so \(\bigcup_{m}A_{m}\in \mathcal{F}\), completing the proof. \(\square\)
Intuitively, \(\sigma(\mathcal{E})\) creates a \(\sigma\)-algebra containing \(\mathcal{E}\) by going from the outside in, that is, starting with too big \(\sigma\)-algebra and going inwards by seeing where they overlap.
For \(\sigma\)-algebra \(\mathcal{F}\), if \(\mathcal{E}\subset \mathcal{F}\) then \(\sigma(\mathcal{E})\subseteq \mathcal{F}\).
PROOF: Since \(\mathcal{F}\) is a \(\sigma\)-algebra containing \(\mathcal{E}\) it must contain the smallest \(\sigma\)-algebra containing \(\mathcal{E}\) since \[ \sigma(\mathcal{E})=\bigcap_{i \in I}\mathcal{F}_{i}, \] where \(\mathcal{F}_{i}\) are the collection of \(\sigma\)-algebra that contain \(\mathcal{E}\). \(\square\)
The following definition is built upon definitions from [topology] which we have included in the [[]].
Let \((X,\tau)\) be a topological space. The Borel σ-algebra of \(X\), \(\mathcal{B}_{X}\) is the \(\sigma\)-algebra generated by \(\tau\). Its members are known as Borel sets.
The Borel \(\sigma\)-algebra of \(\mathbb{R}\) denoted \(\mathcal{B}_{\mathbb{R}}\) is the \(\sigma\)-algebra generated from the standard topology which is the collection of open real intervals.
The Borel \(\sigma\)-algebra of \(\mathbb{R}\), \(\mathcal{B}_{\mathbb{R}}\) is generated by: 1. Open intervals \(\mathcal{E}_{1}:=\{ (a,b):a<b \}\); 2. Closed intervals \(\mathcal{E}_{2}:=\{ [a,b]:a<b \}\); 3. Half-open intervals \(\mathcal{E}_{3}:=\{ (a,b]:a<b \}\); 4. Open rays \(\mathcal{E}_{4}:=\{ (a, \infty):a\in \mathbb{R} \}\); 5. Closed rays \(\mathcal{E}_{5}:=\{ [a,\infty):a\in \mathbb{R} \}\).
PROOF: The Borel \(\sigma\)-algebra \(\mathcal{B}_{\mathbb{R}}:=\sigma(\mathcal{E}_{1})\) is generated from the set of open intervals \[
\mathcal{E}_{1}:=\{ (a,b):a<b \}.
\] (1) Considering arbitrary \((a,b]\in\mathcal{E}_{3}\) we note that since open sets are in \(\mathcal{B}_{\mathbb{R}}\) and \(\sigma\)-algebra are closed under countable intersection we have that \[
(a,b]=\bigcap_{n=1}^\infty \left( a,b+\frac{1}{n} \right)\in\mathcal{B}_{\mathbb{R}}\implies \mathcal{E}_{3}\subseteq\mathcal{B}_{\mathbb{R}\implies}\sigma(\mathcal{E}_{3})\subseteq \mathcal{B}_{\mathbb{R}}.
\] Similarly, considering arbitrary \((a,b)\in\mathcal{E}_{1}\)
\[
(a,b)=\bigcup_{n=1}^\infty \left(a,b-\frac{1}{n}\right]\in \sigma(\mathcal{E}_{3})\implies\mathcal{E}_{1}\subseteq \sigma(\mathcal{E}_{3})\implies\mathcal{B}_{\mathbb{R}}\subseteq \mathcal{E}_{3}.
\] (2) Considering arbitrary \((a,+\infty)\in \mathcal{E}_{5}\) we note that \[
(a,+\infty)=\bigcup_{n=1}^\infty (a,n)\in \mathcal{B}_{\mathbb{R}}\implies\mathcal{E}_{5}\subseteq\mathcal{B}_{\mathbb{R}}\implies \sigma(\mathcal{E}_{5})\subseteq\mathcal{B}_{\mathbb{R}}.
\] Similarly, considering arbitrary \((a,b)\in\mathcal{E}_{1}\) since \(\sigma\)-algebra are closed under countable intersections \[
\begin{align}
(a,b)&=(a,+\infty)\cap[b,+\infty)=(a,+\infty)\cap \bigcap_{n=1}^\infty \left( b-\frac{1}{n},+\infty \right)\in\sigma(\mathcal{E}_{5}) \\
\implies & \mathcal{E}_{1}\subseteq \sigma(\mathcal{E}_{5})\implies \mathcal{B}_{\mathbb{R}}\subseteq \sigma(\mathcal{E}_{5}).
\end{align}
\] (3) Considering arbitrary \([a,+\infty)\in\mathcal{E}_{7}\) we note that \[
\begin{align}
[a,+\infty)&=\bigcap_{n=1}^\infty[a,n)=\bigcap_{n=1}^\infty \bigcap_{m=1}^\infty\left( a-\frac{1}{m}, n \right)\in \mathcal{B}_{\mathbb{R}} \\
\implies & \mathcal{E}_{7}\subseteq \mathcal{B}_{\mathbb{R}}\implies \sigma(\mathcal{E}_{7})\subseteq\mathcal{B}_{\mathbb{R}}.
\end{align}
\] Similarly, considering arbitrary \((a,b)\in \mathcal{E}_{1}\), since \(\sigma\)-algebra are closed under countable unions and complements we have that \[
\begin{align}
(a,b)&=(-\infty,b)\cap(-\infty,a)=([b,+\infty)^c\cup[a,+\infty)^c)^c\subseteq \sigma(\mathcal{E}_{7}) \\
\implies & \mathcal{E}_{1}\subseteq \sigma(\mathcal{E}_{7})\implies\mathcal{B}_{\mathbb{R}}\subseteq \sigma(\mathcal{E}_{7}).
\end{align}
\] \(\square\)
1.2.3 Measure
Given a measurable space \((X,\mathcal{F})\) a measure \(\mu:\mathcal{F}\to[0,+\infty]\) is a mapping frm the collection of measurable sets to \([0,+\infty]\) such that: 1. \(\mu(\emptyset)=0\) 2. \(\{ E_{i} \}_{i=1}^\infty \subseteq \mathcal{F}~\&~\text{disjoint}\implies \mu(\cup_{i=1}^\infty E_{i})=\sum_{i=1}^\infty \mu(E_{i})\).
We call the triple \((X, \mathcal{F}, \mu)\) a measure space. ###### Example (Dirac Measure): For \((X,\mathcal{F})=(X, 2^X)\) we fix \(x_{0}\in X\) and define \[ \mu(A)=\delta_{x_{0}}(A)=\begin{cases} 1&x_{0}\in A \\ 0& \text{o.w.} \end{cases} \] known as the Dirac measure.
For any measure space \((X,\mathcal{F},\mu)\) and \(A,B\in \mathcal{F}\), \(\{ A_{i} \}_{i=1}^n\subseteq \mathcal{F}\) we have the following properties: 1. \(A \subseteq B\implies \mu(A)\leq \mu(B)\) (Monotonicity) 2. \(A \subseteq B\implies \mu(B\setminus A)=\mu(B)-\mu(A)\) 3. \(\mu(\cup_{i=1}^\infty A_{i})\leq \sum_{i=1}^\infty \mu(A_{i})\) (Subadditivity) 4. \(A_{i}\subseteq A_{i+1},~\forall i\implies \mu(\cup_{i=1}^\infty A_{i})=\lim_{_{i \to \infty}}\mu(A_{i})\) (Continuity from below) 5. \(A_{i}\supseteq A_{i+1}~\&~\mu(A_{1})<\infty,~\forall i\implies \mu(\cap_{i=1}^\infty A_{i})=\lim_{_{i \to \infty}}\mu(A_{i})\) (Continuity from above)
PROOF: (1) + (2) Assume there exists \(A\subseteq B\) then we can partition \(B=A\cup(B\setminus A)\) and so \[ \mu(B)=\mu(A)+\mu(B\setminus A)\geq \mu(A). \] (3) Similarly, we can define \(A_{1}=B_{1}\) and \(B_{n}=A_{n}\setminus \cup_{i=1}^{n-1}A_{i}\) then we have that \[ \mu(\cup_{i=1}^\infty A_{i})=\mu(\cup_{i=1}^\infty B_{i})=\sum_{i=1}^\infty \mu(B_{i})=\sum_{i=1}^\infty \mu(A_{i})-\mu(\cup_{i=1}^{n-1}A_{i})\leq \sum_{i=1}^\infty \mu(A_{i}). \] (4) Defining \(B_{1}=A_{1}\) and \(B_{i}=A_{i}\setminus A_{i-1}\) for \(i \geq 2\). By definition \(\cup_{i=1}^n B_{i}=A_{n}\) hence \[ \cup_{i=1}^\infty B_{i}=\cup_{n=1}^\infty A_{n}\implies \mu(A_{n})=\mu(\cup_{i=1}^nB_{i})=\sum_{i=1}^n\mu(B_{i}). \] Consequently taking the limit as \(n \to \infty\) we find \[ \lim_{n \to \infty}\mu(A_{n})=\sum_{i=1}^\infty \mu(B_{i})=\mu\left( \bigcup_{i=1}^\infty B_{i} \right) =\mu\left(\bigcup_{i=1}^n A_{i}\right) \] (5) Define \(B_{i}=A_{1}\setminus A_{i}\). By constraint \(B_{i}\subseteq B_{i+1}\) and so \[ \begin{align} \bigcup_{i=1}^\infty B_{i} & =\bigcup_{i=1}^\infty(A_{1}\setminus A_{i})=\bigcup_{i=1}^\infty(A_{i}\cap A_{i}^c)=A_{1}\cap\left( \bigcup_{i=1}^\infty A_{i}^c \right) \\ & = A_{1}\cap \left( \bigcap_{i=1}^\infty A_{i} \right)^c=A_{1}\setminus \left( \bigcap_{i=1}^\infty A_{i} \right). \end{align} \] Therefore we have that \[ \begin{align} \mu(A_{1}) & =\mu\left( A_{1}\cap\left( \bigcap_{i=1}^\infty A_{i} \right) \right)+\mu\left( A_{1}\setminus \bigcap_{i=1}^\infty A_{i} \right) =\mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\lim_{i \to \infty} \mu(B_{i}) \\ & = \mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\lim_{i \to \infty}\mu(A_{i})-\mu(A_{i} ) = \mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\mu(A_{1})-\lim_{i \to \infty}\mu(A_{i}), \end{align} \] and rearranging gives the result. \(\square\)
Given a measure space \((X,\mathcal{F}, \mu)\) we define the following terminology: 1. the measure is finite if \(\mu(X)<+\infty\) 2. the measure is \(\sigma\)-finite if \(\exists \{ E_{i} \}_{i=1}^\infty \subseteq \mathcal{F}\) such that \(\cup_{i=1}^\infty E_{i}=X\) and \(\mu(E_{i})< \infty\) for all \(i \in \mathbb{N}\) 3. \(E \in S^X\) is a null set (of \(\mu\)) if \(E\in \mathcal{F}\) and \(\mu(E)=0\) 4. a property holds (\(\mu\)-) almost everywhere if the set of points where it fails is a null set 5. a measure whose domain includes all subsets of null sets is called complete.
The next theorem defines a complete measure space, a measure space in which every subset of every null set is measurable (with measure zero).
Suppose that \((X,\mathcal{F}, \mu)\) is a measure space. Let \(\mathcal{N}=\{ N \in \mathcal{F}:\mu(N)=0 \}\) and \(\overline{\mathcal{F}}=\{ E \cup F:E \in\mathcal{F},~F\subset N~\text{for some }N \in\mathcal{N} \}\). Then \(\overline{\mathcal{F}}\) is a \(\sigma\)-algebra known as the completion of \(\mathcal{F}\) with respect to \(\mu\), and there is a unique extension \(\overline\mu\) of \(\mu\) to a complete measure on \(\overline{\mathcal{F}}\), known as the completion of \(\mu\). The resulting measure space \((X,\overline{F}, \overline{\mu})\) is called a complete measure space.
PROOF: Firstly, we show that \(\overline{\mathcal{F}}\) is a \(\sigma\)-algebra. Note that since \(\mathcal{F}\) and \(\mathcal{N}\) are both closed under countable unions, so to is \(\overline{\mathcal{F}}\) and further, since \(\emptyset \in \mathcal{F},\mathcal{N}\) then we have that \(\emptyset \in\overline{\mathcal{F}}\), hence we need only show that \(\overline{\mathcal{F}}\) is closed under compliments.
\(\square\) ## Constructing Measures
Naturally you might ask whether there exists a measure \(\mu\) on the measurable real space \((\mathbb{R}, \mathcal{B}_{\mathbb{R}})\) where \(\mu((a,b))=b-a\) and \(\mu\) is translation invariant. In this section we outline the process whereby we can construct such measures using functions called pre-measures \(\mu_{0}\) and outer-measures \(\mu^*\). We begin by presenting an overview of the process before proceeding to give context and further details in the subsequent sections:
- Start by defining an increasing, right-continuous function \(F\) commonly known as a distribution function.
- From \(F\) we can construct a premeasure \(\mu_{0}\) defined on a small collection of simple sets in \(X\), for example the real intervals for \(X=\mathbb{R}\).
- Extend this premeasure to an outer-measure \(\mu^*\) which by definition is defined on all subsets of \(X\), \(2^X\).
- Use the Caratheodory extension theorem to define the collection of outer-measurable sets \(\mathcal{M}\subset 2^X\).
- Restricting this outer-measure to have domain \(\mathcal{M}\), \(\mu^*|_{\mathcal{M}}\) results in a full measure \(\mu\) on \(\sigma\)-algebra \(\mathcal{M}\).
1.2.4 Outer-Measures
An outer (exterior) measure is a function defined on all subsets of a given set taking values in the extended real numbers satisfying some additional technical conditions
An outer measure on non-empty \(X\) is a function \(\mu^*:2^X\to[0,+\infty]\) s.t.: 1. \(\mu^*(\emptyset)=0\) (null-empty), 2. \(A \subseteq B\implies \mu^*(A)\leq \mu^*(B)\) (monotonic) 3. \(\mu^*\left( \cup_{i=1}^\infty A_{i} \right)\leq \sum_{i=1}^\infty \mu^*(A_{i})\) (countably subadditive).
Note: We often prove (2) and (3) simultaneously since they both hold iff if \(E \subseteq \cup_{i=1}^\infty A_{i}\) then \(\mu^*(E)\leq \sum_{i=1}^\infty \mu^*(A_{i})\).
Given any outer-measure \(\mu^*\), a set \(E \subseteq X\) is (outer) \(\mu^*\)-measurable if \(\forall A \subseteq X\) we have \[ \mu^*(A)=\mu^*(A \cap E)+\mu^*(A \cap E^c). \] We denote the set of outer-measurable sets \(\mathcal{F}_{\mu^*}:=\{ E\subseteq X:E\text{ is }\mu^*\text{-measurable} \}\). Intuitively, we say \(E\) breaks apart any set nicely.
Remark: Note that by countable subadditivity \(\leq\) holds for all \(A \subseteq X\) hence we need only show \(\geq\) for equality.
The most common way of obtaining outer measures is to start with a family \(\mathcal{E}\) of elementary sets on which a notion of measure is defined (such as rectangles in the plane) and then to approximate arbitrary sets “from the outside” by countable unions of members of \(\mathcal{E}\). We state this more precisely in the following proposition.
Let \(\mathcal{E}\subset 2^X\) and \(\mu_{0}:\mathcal{E}\to[0,\infty]\) be such that \(\emptyset \in\mathcal{E}\), \(X\in\mathcal{E}\) and \(\mu_{0}(\emptyset)=0\). For any \(A\subset X\) we can define an outer measure by \[ \mu^*(A)=\inf\left\{ \sum_{j=1}^\infty \mu_{0}(E_{j}):E_{j}\in\mathcal{E},~A\subset \bigcup_{j=1}^\infty E_{j}\right\}.\tag{$\star$} \]
PROOF: For any \(A \subset X\) there exists \(\{ E_{j} \}_{j=1}^\infty \subset\mathcal{E}\) such that \(A\subset \bigcup_{j=1}^\infty E_{j}\) (take \(E_{j}=X\) for all \(j\)) and so the definition of \(\mu^*\) makes sense. We proceed to show this satisfies the properties of outer measure. (1) Firstly note that clearly \(\mu^*(\emptyset)=0\) by taking \(E_{j}=\emptyset\) for all \(j\).
(2) Next for \(A\subset B\) we have that \(\mu^*(A)\leq \mu^*(B)\) since the set over which the infimum is taken in the definition of \(\mu^*(A)\) includes the corresponding set in the definition of \(\mu^*(B)\). (3) To prove countable subadditivity, suppose \(\{ A_{j} \}_{j=1}^\infty \subset2^X\) and fix \(\varepsilon>0\). For each \(j\) there exists \(\{ E_{j}^k \}_{k=1}^\infty \subset\mathcal{E}\) such that \(A_{j}\subset \bigcup_{k=1}^\infty E_{j}^k\) and \(\sum_{k=1}^\infty \mu_{0}(E_{j}^k)\leq \mu^*(A_{j})+\varepsilon{2}^{-j}\). But then, if \(A=\bigcup_{i=1}^\infty A_{j}\), we have \(A\subset \bigcup_{j,k=1}^\infty E_{j}^k\) and \(\sum_{j,k}\mu_{0}(E_{j}^k)\leq \sum_{j}\mu^*(A_{j})+\varepsilon\) whence \(\mu^*(A)\leq \sum_{j}\mu^*(A_{j})+\varepsilon\). Since \(\varepsilon\) is arbitrary, we are done. \(\square\)
Intuitively, we are defining the outer measure of the set \(A\) to be the infimum of all sums of collections of sets that cover \(A\).
For any outer measure \(\mu^*\) if \(\mu^*(B)=0\) then \(B \in \mathcal{F}_{\mu^*}\).
PROOF: We very quickly see that \[ \mu^*(E)\geq \mu^*\left( E \cap B\right)+\mu ^*(E \cap B^c)=\mu^*(E\cap B^c). \] \(\square\)
1.2.5 Pre-measures
A pre-measure is a set function that acts as a pre-cursor to a bona fide measure through the Carathéodory Extension Theorem.
For algebra \(\mathcal{A}\subset 2^X\) then a function \(\mu_{0}:\mathcal{A}\to[0,\infty]\) is called a premeasure if is satisfies both: 1. \(\mu_{0}(\emptyset)=0\); and 2. For disjoint \(\{ A_{j} \}_{i=1}^\infty \subset\mathcal{A}\) such that \(\cup_{i=1}^\infty A_{i}\in \mathcal{A}\) then \(\mu_{0}(\cup_{i=1}^\infty A_{i})=\sum_{i=1}^\infty \mu_{0}(A_{i})\).
In particular, a premeasure is finitely additive since one can take \(A_{j}=\emptyset\) for large \(j\).
If \(\mu_{0}\) is a premeasure on \(\mathcal{A}\) and \(\mu^*\) is given by \((\star)\) as \[ \mu^*(A)=\inf\left\{ \sum_{j=1}^\infty \mu_{0}(E_{j}):E_{j}\in\mathcal{E},~A\subset \bigcup_{j=1}^\infty E_{j}\right\}, \] then: 1. \(\mu^*|_{\mathcal{A}}=\mu_{0}\); and 2. Every set \(A\in\mathcal{A}\) is \(\mu^*\)-measurable.
PROOF: (1) Letting \(E \in\mathcal{A}\), if \(E \subset \bigcup_{j=1}^\infty A_{j}\) with \(A_{j}\in\mathcal{A}\) we define \[ B_{n}=E \cap \left( A_{n}\setminus\bigcup_{j=1}^{n-1}A_{j} \right). \] Then \(B_{n}\) are disjoint members of \(\mathcal{A}\) whose union is \(E\) thus \[ \mu_{0}(E)=\sum_{j=1}^{\infty}\mu_{0}(B_{j})\leq \sum_{j=1}^{\infty}\mu_{0}(A_{j}). \] It follows that \(\mu_{0}(E)\leq \mu^*(E)\) and the reverse inequality is clear since \(E\subset \bigcup_{j=1}^\infty A_{j}\) where \(A_{1}=E\) and \(A_{j}=\emptyset\) for \(j>1\). (2) If \(A \in\mathcal{A}\), \(E \subset X\) and \(\epsilon>0\), there is a sequence \(\{ B_{j} \}_{j=1}^\infty \subset \mathcal{A} \mathcal{}\) with \(E \subset \bigcup_{j=1}^\infty B_{j}\) and \(\sum_{j=1}^{\infty}\mu_{0}(B_{j})\leq \mu^*(E)+\epsilon\). Since \(\mu_{0}\) is additive on \(\mathcal{A}\) we have \[ \begin{align} \mu^*(E)+\epsilon & \geq \sum_{j=1}^{\infty}\mu_{0}(B_{j}\cap A) + \sum_{j=1}^{\infty} \mu_{0}(B_{j}\cap A^c) \\ & \geq \mu^*(E \cap A) + \mu^*(E \cap A^c). \end{align} \] Since \(\epsilon\) is arbitrary, \(A\) is \(\mu^*\)-measurable. \(\square\)
1.2.6 Carathéodory Extension Theorem
The next key theory from Carathéodory shows that if we are given an outer-measure \(\mu^*\) then the set of outer-measurable sets \(\mathcal{F}_{\mu^*}\) is a \(\sigma\)-algebra and when restricted to this \(\sigma\)-algebra, \(\mu^*\) becomes a measure.
Given and outer measure \(\mu^*\) then 1. \(\mathcal{F}_{\mu^*}\) is a \(\sigma\)-algebra; and 2. \(\mu^*\) is a measure on \(\mathcal{F}_{\mu^*}\).
Note: We can show that this is in general not the largest \(\sigma\)-algebra on which \(\mu^*\) is a measure.
PROOF: Since \(\mu^*(\emptyset)=0\implies \emptyset \in \mathcal{F}_{\mu^*}\) by the proposition above, hence \(\mathcal{F}_{\mu^*}\) is non-empty; and by definition of an outer-measurable set \(\mathcal{F}_{\mu^*}\) is closed under compliments. Next, if \(A,B \in \mathcal{F}_{\mu^*}\) and \(E\subset X\) we have that \[ \begin{align} \mu^*(E) & = \mu^*(E \cap A)+\mu^*(E \cap A^c) \\ & = \mu^*(E \cap A \cap B)+\mu^*(E \cap A \cap B^c)+\mu^*(E \cap A^c \cap B)+\mu^*(E \cap A^c \cap B^c) \\ & \geq \mu^*(E \cap(A\cup B))+\mu^*(E \cap(A \cup B)^c), \end{align} \] where the final inequality holds from subadditivity since \(A\cup B=(A \cap B)\cup(A \cap B^c)\cup (A^c\cap B)\). It follows that \(A \cup B\in \mathcal{F}_{\mu^*}\) and so it is an algebra. Moreover, if \(A,B \in \mathcal{F}_{\mu^*}\) and \(A \cap B=\emptyset\) then \[ \mu^*(A \cup B)=\mu^*((A\cup B)\cap A)+\mu^*((A \cup B)\cap A^c)=\mu^*(A)+\mu^*(B), \] hence \(\mu^*\) is finitely additive on \(\mathcal{F}_{\mu^*}\). To prove \(\mathcal{F}_{\mu^*}\) is a \(\sigma\)-algebra it suffices to show \(\mathcal{F}_{\mu^*}\) is closed under countable disjoint unions. If we consider disjoint \(\{ A_{j} \}_{j=1}^\infty \subseteq \mathcal{F}_{\mu^*}\) and define \(B_{n}:=\bigcup_{j=1}^nA_{{j}}\) and \(B :=\bigcup_{j=1}^\infty A_{j}\). Then, for any \(E \subset X\) we have \[ \mu^*(E \cap B_{n})=\mu^*(E \cap B_{n}\cap A_{n})+\mu^*(E \cap B_{n}\cap A_{n}^c)=\mu^*(E \cap A_{n})+\mu^*(E \cap B_{n-1}), \] hence, by induction we have that \(\mu^*(E \cap B_{n})=\sum_{j=1}^n\mu^*(E \cap A_{j})\). Therefore \[ \begin{align} \mu^*(E) & =\mu^*(E \cap B_{n})+\mu^*(E \cap B_{n}^c)\geq \sum_{j=1}^n\mu^*(E\cap A_{j})+\mu^*(E \cap B^c) \\ & \stackrel{n \to \infty}{\geq }\sum_{j=1}^\infty \mu^*(E \cap A_{j})+\mu^*(E \cap B^c)\geq \mu^*\left( \bigcup_{j=1}^\infty (E\cap A_{j}) \right)+\mu^*(E \cap B^c) \\ & = \mu^*(E \cap B)+\mu^*(E\cap B^c)\geq \mu^*(E), \end{align} \] and so all the inequalities become equalities thus \(B\in \mathcal{F}_{\mu^*}\) as required. Furthermore, taking \(E=B\) we have \(\mu^*(B)=\sum_{j=1}^\infty \mu^*(A_{j})\) and so \(\mu^*\) is countable additive on \(\mathcal{F}_{\mu^*}\). Finally, if \(\mu^*(A)=0\), for any \(E \subset X\) \[ \mu^*(E)\leq \mu^*(E \cap A)+\mu^*(E \cap A^c)=\mu^*(E \cap A^c)\leq \mu^*(E), \] so that \(A \in \mathcal{F}_{\mu^*}\), hence \(\mu^*\) is a complete measure on \(\mathcal{F}_{\mu^*}\). \(\square\)
1.2.7 Lebesgue Measures
Now let us turn our focus to the real measurable space \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) and the most important measure in Real Analysis, namely the Lebesgue measure which is a type of Borel measure. A Borel measure is any measure defined on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). To motivate the following section, if we consider a finite Borel measure \(\mu\) and define the distribution function \[ F(x)=\mu((-\infty,x]) \] we note that by construction: (1) \(F\) is increasing; and (2) \(F\) is right-continuous. Our aim is to invert this process, namely to start with an increasing and right-continuous function and use it to construct a measure through the steps outlines at the start of [[]].
Sets of the form \((a,b]\), \((a,\infty)\) or \(\emptyset\) for \(-\infty \leq a<b<\infty\) are known as \(h\)-intervals . The collection \(\mathcal{A}\) of finite disjoint unions of \(h\)-intervals is an algebra and \(\sigma(\mathcal{A})=\mathcal{B}_{\mathbb{R}}\).
For increasing and right-continuous \(F:\mathbb{R} \to \mathbb{R}\) for disjoint \(h\)-intervals \((a_{j},b_{j}]\) for \(j=1, \dots, n\) the function defined by \[ \begin{cases}\mu_{0}\left(\bigcup_{j=1}^n (a_{j}, b_{j}]\right)=\sum_{j=1}^{n} [F(b_{j})-F(a_{j})] \\ \mu_{0}(\emptyset)=0\end{cases} \] is a premeasure on the algebra \(\mathcal{{A}}\).
An important outer-measure in real analysis and Stochastic Process is the Lebesgue outer-measure \(\lambda^*\).
The Lebesgue outer measure \(\lambda^*:2^\mathbb{R}\to [0,+\infty]\) is defined \[ \lambda^*(A)=\inf\left\{ \sum_{i=1}^\infty \lvert b_{i}-a_{i} \rvert :A \subseteq \bigcup_{i=1}^\infty (a_{i},b_{i}] \right\}. \]
Clearly from the previous result the function is an outer measure. We can also show: (1) \(\lambda^*((a,b])=b-a\), \(a\leq b\); and (2) \(\lambda^*\) is translation invariant. Importantly, it is not countable additive however, we will how that it becomes countably additive when restricted to “nice enough” sets.
1.2.8 Borel Measures
PROOF:
If \(F:\mathbb{R}\to \mathbb{R}\) is any increasing, right-continuous function, there is a unique Borel measure \(\mu_{F}\) on \(\mathbb{R}\) such that for all \(a,b\) we have \[ \mu_{F}((a,b])=F(b)-F(a). \] If \(G\) is another such function, we have \(\mu_{F}=\mu_{G}\) iff \(F-G\) is constant. Conversely, if \(\mu\) is a Borel measure on \(\mathbb{R}\) that is finite on all bounded Borel sets and we define \[ F(x)=\begin{cases}\mu((0,x]) & x>0\\ 0 & x=0 \\ -\mu((-x,0]) & x<0\end{cases}, \] then \(F\) is increasing and right-continuous, and \(\mu=\mu_{F}\).
The completion \(\overline{\mu}_{F}\) (which is often also denoted \(\mu_{F}\)) is called the Lebesgue-Stieltjes measure associated to \(F\).
PROOF:
The theory of this section gives, for each increasing and right continuous \(F\), not only the Borel measure \(\mu_{F}\) but a complete measure \(\overline{\mu}_{F}\) whose domain includes \(\mathcal{B}_{\mathbb{R}}\). In fact, \(\overline{\mu}_{F}\) is just the completion of \(\mu_{F}\) and one can show that its domain is always strictly larger than \(\mathcal{B}_{\mathbb{R}}\). We shall usually denote this complete measure also by \(\mu_{F}\) and call it the Lebesgue-Stieltjes measure associated to \(F\).
Fix a complete Lebesgue-Stieltjes measure \(\mu\) on \(\mathbb{R}\) associated to the increasing, right continuous function \(F\) and denote by \(\mathcal{M}_{\mu}\) the domain of \(\mu\).
For any \(E \in \mathcal{M}_{\mu}\) \[ \mu(E)=\inf_{}{\left\{ \sum_{j=1}^{\infty}\mu((a_{j},b_{j})):E\subset \bigcup_{j=1}^\infty (a_{j},b_{j}) \right\}}. \]
PROOF:
If \(E \in \mathcal{M}_{\mu}\) then \[ \begin{align}\mu(E) & =\inf_{}{\{ \mu(U):E\supset E,~E\text{ is open} \}} \\ & = \sup_{}{\{ \mu(K):K \subset E,~K \text{ is compact} \}}.\end{align} \]
PROOF:
If \(E \subset \mathbb{R}\) the following are equivalent: 1. \(E \in \mathcal{M}_{\mu}\), 2. \(E=V-N_{1}\) where \(V\) is a \(G_{\delta}\) set and \(\mu(N_{1})=0\), 3. \(E=H \cup N_{2}\) where \(H\) is a \(F_{\sigma}\) set and \(\mu(N_{2})=0\).
PROOF:
If \(E \in \mathcal{M}_{\mu}\) and \(\mu(E)<\infty\) then for every \(\epsilon>0\) there is a set \(A\) that is a finite union of open intervals such that \(\mu(E \triangle A)<\epsilon\).
PROOF:
The complete measure \(\mu_{F}\) associated to the function \(F(x)=x\) is the Lebesgue measure which we denote by \(\lambda\). The domain of \(\lambda\) is the class of Lebesgue measurable sets which we denote \(\mathcal{L}\).
We note that we shall also refer to the restriction of \(\lambda\) to \(\mathcal{B}_{\mathbb{R}}\) as the Lebesgue measure.
If \(E\in \mathcal{L}\) then \(E+s \in \mathcal{L}\) and \(rE \in \mathcal{L}\) for all \(s,r \in \mathbb{R}\). Moreover \(\lambda(E+s)=\lambda(E)\) and \(\lambda(rE) =\lvert r \rvert \lambda(E)\).
PROOF:
1.3 Lebesgue Integrals
1.3.1 Measurable Functions
We recall that any mapping \(f:X \to Y\) between two sets induces a mapping \(f^{-1}:2^Y\to 2^X\) defined by \[ f^{-1}(E)=\{ x \in X:f(x)\in E \}, \] which preserves unions, intersections and complements, i.e. \[ \begin{align} f^{-1}\left( \bigcup_{{\alpha}}^{}{E_{\alpha}} \right) & =\bigcup_{\alpha}f^{-1}(E_{\alpha}) \\ f^{-1}(E^c) & =(f^{-1}(E))^c \\ f^{-1}\left( \bigcap_{\alpha}E_{\alpha} \right) & =\bigcap_{\alpha}f^{-1}(E_{\alpha}). \end{align} \] Supposing \((X,\mathcal{M})\) and \((Y,\mathcal{N})\) are measurable spaces then the pull-back and push-forward of \(\mathcal{N}\) respectively defined as \[ \begin{align} \{ f^{-1}(E):E\in \mathcal{N} \} \quad \&\quad\{ E:f^{-1}(E)\in \mathcal{N} \}, \end{align} \] are \(\sigma\)-algebra.
For measurable spaces \((X,\mathcal{M})\) and \((Y,\mathcal{N})\), a mapping \(f:X \to Y\) is said to be \((\mathcal{M}, \mathcal{N})\)-measurable if $$ \[\begin{align}\forall E \in\mathcal{N},&~~f^{-1}(E):=\{ E \in X:f(x) \in E \}\in\mathcal{M}, \\ ::: \iff & \{ \text{The preimage of every measurable set is measurable.} \}\\ \iff& \{ f^{-1}(E):E \in \mathcal{N} \}\subset \mathcal{M}, \\ \iff& \{ E:f^{-1}(E) \in \mathcal{M} \}\supset \mathcal{N}.\end{align}\] $$
Note: This is very closely related to the definition of continuity of [[topological-functions]], i.e. for topological spaces \(X\) and \(Y\), every continuous function \(f:X \to Y\) is Borel measurable.
We say that a real function \(f:\mathbb{R}\to \mathbb{R}\) is Lebesgue measurable if it is \((\mathcal{F}_{\lambda^*},\mathcal{B}_{\mathbb{R}})\)-measurable. Also, given topological spaces \(X,Y\) we say that \(f:X \to Y\) is Borel measurable if it is \((\mathcal{B}_{X},\mathcal{B}_{Y})\)-measurable. We will see that for functions \(f:\mathbb{R} \to \mathbb{R}\) we have that Borel measurability implies Lebesgue measurability since \(\mathcal{B}_{\mathbb{R}}\subsetneq \mathcal{F}_{\lambda^*}\).
Given measurable spaces \((X,\mathcal{M})\) and \((Y,\mathcal{N})\) where \(\mathcal{N}\) is generated by \(\mathcal{E}\), then \(f:X \to Y\) is \((\mathcal{N},\mathcal{M})\)-measurable if and only if \(\forall E\in \mathcal{E}\), \(f^{-1}(E)\in\mathcal{N}\).
PROOF: The only if implication is trivial. For the converse we observe that \[ \{ E \subset Y:f^{-1}(E)\in \mathcal{F} \}, \] is a \(\sigma\)-algebra that contains \(\mathcal{E}\); it therefore contains \(\mathcal{G}\). \(\square\)
For measurable space \((X,\mathcal{F})\) and function \(f:X \to \mathbb{R}\) the following are equivalent: 1. \(f\) is \(\mathcal{F}\)-measurable; 2. \(f^{-1}((a,\infty))\in \mathcal{F}\) for all \(a\in \mathbb{R}\); 3. \(f^{-1}([a,\infty))\in \mathcal{F}\) for all \(a \in \mathbb{R}\); 4. \(f^{-1}((-\infty,a))\in \mathcal{F}\) for all \(a \in \mathbb{R}\); 5. \(f^{-1}((-\infty,a])\in \mathcal{F}\) for all \(a \in \mathbb{R}\).
PROOF: We have previously shown that the open and closed rays generate the Borel σ-Algebra. Thus from previous results we have the result. \(\square\) ###### Example (Measurability of constant real functions): We show that constant functions are always measurable, that is \(f:X \to \overline{\mathbb{R}}\) given by \(f(x)=c\) for all \(x \in X\) for \(c\in \overline{\mathbb{R}}\) if \(f\) is measurable? For any \(E \in \mathcal{{B}}_{\mathbb{R}}\)
\[
f^{-1}(E)=\begin{cases}
\emptyset & c \not\in E, \\
X & c \in E.
\end{cases}
\] Thus, any \(\sigma\)-algebra \(\mathcal{M}\) contains \(\emptyset\) and \(X\). ###### Example (Measurability of indicator functions): We show that indicator functions \(\mathbb{1}_{}:X \to \mathbb{R}\) are always measurable. For function \[
\mathbb{1}_{E}(x):=\begin{cases}
1 & x \in E, \\
0 & x \not\in E,
\end{cases}
\] with measurable \(E\), taking intervals \((a,b]\in \mathbb{R}\) \[
\mathbb{1}_{E}^{-1}((a,b])=\begin{cases}
X & 0,1 \in(a,b], \\
X-E & 0 \in(a,b], \\
E & 1 \in(a,b], \\
\emptyset & 0,1 \not\in (a,b].
\end{cases}
\] Since these sets are all measurable the function is measurable.
For measurable functions \(f,g:X \to \mathbb{R}\), then the following are also measurable: 1. \(f+g\) (with convention \(+\infty+(-\infty)=0\)), 2. \(fg\) (with convention \(0\cdot\pm \infty=0\)), 3. \(f \wedge g\), 4. \(f \vee g\), 5. \(f\circ g\).
PROOF:
For sequences of measurable function \(f_{i};X \to \mathbb{R}\) for \(i \in \mathbb{N}\) then the following are also measurable: 1. \(\sup_{i}{f_{i}}\), 2. \(\inf_{i}{f_{i}}\), 3. \(\limsup_{i}{f_{i}}\), 4. \(\liminf_{i}{f_{i}}\), 5. \(\lim_{ i \to \infty }f_{i}\) as long as the limit exists pointwise.
PROOF: