1 Infimum and Supremum
1.1 Extended Real Sequences
1.1.1 Boundedness of Real Sequences
Consider the extended real sequence \(\{ x_{n} \}_{n \in \mathbb{N}}\subseteq\overline{\mathbb{R}}=:\mathbb{R}\cup \{ \pm \infty \}\). The number \(u \in \mathbb{R}\) is said to be an upper bound of the sequence if \(x_{n}\leq u\) for all \(n \in \mathbb{N}\). We say that the sequence is bounded above. Similarly, the number \(l \in \overline{\mathbb{R}}\) is said to be lower bound of the sequence if \(x_{n}\geq l\) for all \(n \in \mathbb{N}\). We say that the sequence is bounded below. A sequence is bounded if it is both bounded above and below. ### Supremum and Infimum of Real Sequences
The supremum \(\sup_{n\in \mathbb{N}}{x_{n}}\) of extended real sequence \(x_{n}\) is the unique least upper bound, that is it satisfies 1. \(\sup_{n \in \mathbb{N}}{x_{n}}\) is an upper bound of sequence \(x_{n}\); and 2. For set of upper bounds \(U\) we have that \(\sup_{n \in \mathbb{N}}x_{n}\leq u\) for all \(u \in U\).
The infimum \(\sup_{n\in \mathbb{N}}{x_{n}}\) of extended real sequence \(x_{n}\) is the unique greatest lower bound, that is it satisfies 1. \(\inf_{n \in \mathbb{N}}{x_{n}}\) is a lower bound of sequence \(x_{n}\); and 2. For set of lower bounds \(L\) we have that \(\inf_{n \in \mathbb{N}}x_{n}\geq l\) for all \(l \in L\).
These are fundamental definitions in Measure Theory Notes and Real Analysis where they form the bases for defining limits, continuity and integrals.
For a extended real sequence \(\{ x_{n} \}_{n \in \mathbb{N}}\) we have that \[ \inf_{n \in \mathbb{N}}{x_{n}}\leq x_{n}\leq \sup_{n \in \mathbb{N}}{x_{n}};~~\forall n. \]
PROOF: Assuming the result is false then \(\inf_{}x_{n}>\sup{x_{n}}\). However since the infimum is a lower bound and the supremum is an upper bound we have \(\inf_{}{x_{n}}\leq x_{n}\) and \(\sup{x_{n}}\geq x_{n}{}\) for all \(n \in \mathbb{N}\), thus \[ \inf_{}{x_{n}}\leq x_{n}\leq \sup_{}{x_{n}}<\inf_{}{x_{n}}, \] which is a contradiction \(\rightarrow\leftarrow\). \(\square\)
For sequences where \(x_{n}\leq y_{n}\) for all \(n\) we have \[ \begin{align}\inf_{}{x_{n}}&\leq \inf_{}{y_{n}}\\ \sup_{}{x_{n}}&\leq \sup_{}{y_{n}}.\end{align} \]
PROOF: Suppose this is not the case. Let \(m\) be index such that \(y_{m}=\inf_{}{y_{n}}\). Then \[ \inf_{}{x_{n}}\leq x_{m}\leq y_{m}=\inf_{}{y_{n}}, \] which is a contradiction \(\rightarrow\leftarrow\). A similar result holds for the supremum. \(\square\)
For sequences \(\{ x_{n} \},\{ y_{n} \}\) we have \[ \begin{align}\sup(x_{n}+y_{n}) & \leq \sup_{}{x_{n}}+\sup_{}{y_{n}} \\ \inf_{}{(x_{n}+y_{n})} & \geq \inf_{}{x_{n}}+\inf_{}{y_{n}}.\end{align} \]
PROOF: Let \(m\) be index such that \(\sup_{}{(x_{n}+y_{n})}=x_{m}+y_{m}\) then \[ \sup_{}{(x_{n}+y_{n})}= x_{m}+y_{m}\leq \sup_{}{x_{n}}+\sup_{}{y_{n}}. \] Similarly, let \(m\) be index such that \(\inf_{}{(x_{n}+y_{n})}=x_{m}+y_{m}\) then \[ \inf_{}{(x_{n}+y_{n})}=x_{m}+y_{m}\geq \inf_{}{x_{n}}+\inf_{}{y_{n}}. \] \(\square\)
For sequences \(\{ x_{n} \}, \{ y_{n} \}\) we have \[ \begin{align}\sup_{}{(x_{n}\cdot y_{n})}\leq \sup_{}{x_{n}}\cdot \sup_{}{y_{n}}\\ \inf_{}{(x_{n}\cdot y_{n})}\geq \inf_{}{x_{n}}\cdot \inf_{}{y_{n}}.\end{align} \]
PROOF: The proof is similar to the proposition of inequalities of sums. \(\square\)
1.2 Sets
1.2.1 Supremum and Infimum of Sets
1.3 Limit Supremum and Limit Infimum
The limit supremum \(\limsup_{n \to \infty}{x_{n}}\) of extended real sequence \(x_{n}\) is defined as \[ \limsup_{n \to \infty}{x_{n}}=\lim_{ n \to \infty } \sup_{k \geq n}{x_{k}}. \] The limit infimum \(\liminf_{n \to \infty}{x_{n}}\) of extended real sequence \(x_{n}\) is defined as \[ \liminf_{n \to \infty}{x_{n}}=\lim_{ n \to \infty } \inf_{k \geq n}{x_{k}}. \] Intuitively, we are considering the supremum and infimum of the tail \(\{ x_{k} \}_{k \geq n}\) for each step \(n\) and looking at the limit of the supremum and infimum as \(n \to \infty\). They are the largest and smallest numbers respectively that the sequence gets arbitrarily close to infinitely often.
The limit supremum and limit infimum always exist for extended real sequences.
Since the sequences \(\{ \sup_{k \geq n}{x_{k}} \}_{n\in \mathbb{N}}\) and \(\{ \inf_{k \geq n}{x_k} \}_{n \in \mathbb{N}}\) are non-increasing and non-decreasing respectively in \(n\) we can introduce the following extension to the definitions above \[ \begin{align} \limsup_{n \to \infty}{x_{n}} & =\inf_{n\in \mathbb{N}}\sup_{k \geq n}{x_{k}} \\ \liminf_{n \to \infty}{x_{n}} & =\sup_{n\in \mathbb{N}}\inf_{k \geq n}{x_{k}}. \end{align} \]
For an extended real sequence \(\{ x_{n} \}\subseteq\overline{\mathbb{R}}\) the following inequality always holds \[ \liminf_{n \to \infty}{x_{n}}\leq \limsup_{n \to \infty}{x_{n}}. \]
PROOF: We define \(i_{n}=\inf_{k \geq n}{x_{k}}\) and \(s_{n}=\sup_{k \geq n}{x_{k}}\). Since \(\{ i_{n} \}\) is increasing we have that \(\lim_{ n \to \infty }i_{n}=\sup_{n}{i_{n}}\). Similarly, since \(\{ s_{n} \}\) is decreasing we have that \(\lim_{ n \to \infty }s_{n}=\inf_{n}{s_{n}}\) and from [[|previous results]] we have that \(\lim_{ n \to \infty }i_n=\sup_{n}{i_{n}}\leq \inf_{n}{s_{n}}=\lim_{ n \to \infty }s_{n}\), which gives the result. \(\square\)
For an extended real sequence \(\{ x_{n} \}\subseteq\overline{\mathbb{R}}\) assume that \(\liminf_{}{x_{n}}=\limsup_{}{x_{n}}\). Then the limit \(\lim_{ n \to \infty }x_{n}\) exists in \(\overline{\mathbb{R}}\) and \[ \limsup_{}{x_{n}}=\liminf_{}{x_{n}}=\lim_{ }x_{n}. \]
PROOF:
For extended real sequences \(\{ x_{n} \}, \{ y_{n} \}\) where one is dominated by the other, that is \(x_{n}\leq y_{n}\) for all \(n\) then \[ \begin{align}\liminf_{}{x_{n}}\leq \liminf_{}{y_{n}} \\ \limsup_{}{x_{n}}\leq \limsup_{}{y_{n}}\end{align}. \]
PROOF: For all \(n\) defining \(m_{n}\) such that \(y_{m_{n}}:=\inf_{k\geq n}{y_{n}}\) we have that \[ i_{n}^x:=\inf_{k \geq n}{x_{k}}\leq x_{m_{n}}\leq y_{m_{n}}=\inf_{k \geq n}{y_{n}}=:i_{n}^y. \] Similarly, defining \(k_{n}\) such that \(\sup_{k \geq n}i_{n}^x=:i_{k_{n}}\) we have that \[ \sup_{k \geq n}i_{n}^x=i_{k_{n}}^x\leq i_{k_{n}}^y\leq \sup_{k \geq n}{i_{n}^y}, \] giving the first result. A similar procedure gives the second result. \(\square\)
For extended real sequences \(\{ x_{n} \}, \{ y_{n} \}\) the following inequalities hold \[ \begin{align}\limsup_{}{(x_{n}+y_{n})}\leq \limsup_{}{x_{n}}+\limsup_{}{y_{n}} \\ \liminf{(x_{n}+y_{n})}\geq \liminf_{}{x_{n}}+\liminf_{}{y_{n}}.\end{align} \]
PROOF:
For extended real sequence \(\{ x_{n} \}\) we have that \[ \limsup_{}{-x_{n}}=-\liminf_{}{x_{n}}. \]
PROOF:
1.4 Sequences of Sets
Consider a sequence of sets \(\{ A_{n} \}_{n\in \mathbb{N}}\). In the context of the limsup and liminf of sets, the supremum of the sequence \(\{ A_{n} \}_{n\in \mathbb{N}}\) is the smallest possible upper bound for every element (set) of the sequence, i.e the smallest set containing each \(A_{n}\) which by definition is \[ \sup_{n \geq1}A_n= \bigcup_{n=1}^\infty A_n. \] Similarly, the infimum of the sequence \(\{ A_{n} \}_{n\in\mathbb{N}}\) is the largest possible lower bound of every element (set) of the sequence, i.e. the smallest collection of elements in each \(A_{n}\) which by definition is \[ \inf_{n \geq 1} A_{n}=\bigcap_{n=1}^\infty A_n. \]
1.4.1 Set Theoretic Limit Supremum & Infimum
Generally the limit supremum and limit infimum consider what happens to the supremum and the infimum as you disregard an increasingly larger finite collection of sets at the start of the sequence. So for the supremum we consider the union as we increase the lower bound \[ \limsup_{ n \to \infty } A_{n}=\lim_{ n \to \infty } \bigcup_{m=n}^\infty A_{m}. \] However, since the sequence \(\{ B_{n} \}_{n \in \mathbb{N}}\) where \(B_{n}=\bigcup_{m=n}^\infty A_{m}\) is monotone decreasing, this limit is equivalent to the intersection and so \[ \limsup_{ n \to \infty } A_{n}=\lim_{ n \to \infty } \bigcup_{m=n}^\infty A_{m}=\bigcap_{n=1}^\infty \bigcup_{m=n}^\infty A_{m}. \] Similarly, for the infimum we consider the intersection as we increase the lower bound \[ \liminf_{ n \to \infty } A_{n}=\lim_{ n \to \infty } \bigcap_{m=n}^\infty A_{m}. \] Again, since the sequence \(\{ B_{n} \}_{n \in \mathbb{N}}\) where \(B_{n}:=\bigcap_{m=n}^\infty A_{m}\) is monotone increasing, this limit is equivalent to the unions as so \[ \liminf_{ n \to \infty } A_{n} = \lim_{ n \to \infty } \bigcap_{m=n}^\infty A_{m}=\bigcup_{n=1}^\infty \bigcap _{m=n}^\infty A_{m}. \]
1.5 Probability Theory
1.5.1 Limsup and Liminf of Probability
An important notion in probability theory, specifically in the convergence of random variables and the Borel-Cantelli lemma is the notion of events occurring infinitely often which, somewhat counterintuitively, is equivalent to the notion of even limit supremum, i.e. for sequence of events \(A_{n}\), we have that \[ \mathbb{P}(\{ A_{n}\text{ occurs }i.o \}) = \mathbb{P}\left(\limsup_{ n \to \infty } A_{n}\right). \] A powerful tool for constructing proofs that events occur i.o. is the following theorem.
For a sequence of events \(\{ A_{n} \}_{n \in \mathbb{N}}\) we have that \[ \mathbb{P}\left(\liminf_{ n \to \infty } A_{n}\right) \leq \liminf_{ n \to \infty } \mathbb{P}(A_{n})\leq \limsup_{ n \to \infty }\mathbb{P}(A_{n})\leq \mathbb{P}\left( \limsup_{ n \to \infty } A_{n} \right). \]
PROOF: The second inequality is clear by definition hence we show the first (i) and third (iii) inequalities. For (iii) we have that \[
\begin{align}
\limsup_{ n \to \infty } \mathbb{P}(A_{n}) & =\inf_{n\geq 1} \sup_{m \geq n}\mathbb{P}(A_{m})\stackrel{(\star)}{\leq} \inf_{n\geq 1}\sup_{m \geq n}\mathbb{P}\left( \bigcup_{k=m}^\infty A_{k} \right) \\
& =\inf_{n \geq 1}\mathbb{P}\left( \bigcup _{k=n}^\infty A_{k} \right)=\mathbb{P}\left( \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_{k} \right)=\mathbb{P}\left( \limsup_{ n \to \infty }A_{n} \right) ,
\end{align}
\] where for \((\star)\) we used the monotonicity of measure and making use of the fact that \(\bigcup_{k=n}^\infty A_{k}\downarrow\) as \(n\uparrow\). For (i) we simply replace \(A_{n}\) with its complement \[
\limsup_{ n \to \infty } \mathbb{P}(A_{n}^c) \leq \mathbb{P}\left( \limsup_{ n \to \infty } A_{n}^c \right),
\] and then we obtain the result since \[
\mathbb{P}\left( \liminf_{ n \to \infty } A_{n} \right) =1-\mathbb{P}\left( \limsup_{ n \to \infty } A_{n}^c \right) \leq 1-\limsup_{ n \to \infty } \mathbb{P}(A_{n}^c)=\liminf_{ n \to \infty } (1-\mathbb{P}(A_{n}^c))=\liminf_{ n \to \infty } \mathbb{P}(A_{n}).
\] \end{proof}