Riemann Integral
1 Riemann Integration
Let \(f:[a,b]\to \mathbb{R}\) be a bounded (not necessarily continuous) function on a compact (closed, bounded) interval. We will define what it means for \(f\) to be Riemann integrable on \([a,b]\) and, in that case, define its Riemann integral \(\int_a^b f\). The integral of \(f\) on \([a,b]\) is a real number whose geometrical interpretation is the signed area under the graph \(y=f(x)\) for \(a\leq x\leq b\).
The Riemann integral is the simplest integral to define and it allows one to integrate every continuous function as well as some not-too-badly discontinuous functions. For highly discontinuous functions however it is insufficient and we instead turn to the Lebesgue Integral which is more flexible with better mathematical properties.
1.1 Riemann Integral
We say that two intervals are almost disjoint if they are disjoint or intersect only at a common endpoint.
Let \(I\) be a nonempty compact interval. A partition of \(I\) is a finite collection \(\{I_1, ..., I_n\}\) of almost disjoint, nonempty, compact subintervals whose union is \(I\).
We denote a partition of \([a,b]\) by with subintervals \(I_k=[x_{k-1},x_k]\) either by the collection of intervals \[ \mathcal{P}=\{I_1, ..., I_n\}, \] or by the set of endpoints \[ \mathcal{P}=\{a=x_0 < x_1 < \cdots< x_n = b\}. \] Note that the sum of lengths \(|I_k|=x_k-x_{k-1}\) of the almost disjoint subintervals in a partition of an interval is equal to the length of the whole interval. Suppose that \(f:[a,b]\to\mathbb{R}\) is a bounded function on the compact interval \(I=[a,b]\) with \[ M=\sup_I f\quad \& \quad m=\inf_I f. \] If \(\mathcal{P}\) is a partition of \(I\), let \[ M_k=\sup_{I_k}f\quad \& \quad m_k=\inf_{I_k}f. \] Note that these suprema and infima are well-defined, finite real numbers since \(f\) is bounded. Moreover \[ m\leq m_k\leq M_k<M. \] If \(f\) is continuous on the interval \(I\), then it is bounded and attains its maximum and minimum values on each subinterval, but a bounded discontinuous function need not attain its supremum or infimum.
Further, we define the mesh or net of the partition by \[ ||\mathcal{P}||=\max_i(x_i-x_{i-1}). \] We say that \(f\) is Riemann integrable on \([a,b]\) if there exists a real number \(I\) such that for all sample points \(\xi_i\in[x_{i-1}, x_i]\) \[ \lim_{||\mathcal{P}||\to 0}\sum_{i=1}^n f(\xi_i)(x_i - x_{i-1})=I, \] that is, as we make the mesh finer and finer this sum (of rectangles) approaches some constant. That common limit (if it exists) is denoted by \[ \int_a^b f(x)~dx. \]
1.2 Riemann Integrability
Riemann integrals are the first type of integrals we are introduced to in our study of calculus due to their ease of interpretability and their utility for application to smooth and continuous functions. However, it is not too challenging to define a function that is not Riemann integrable, for example
A bounded function on a compact interval \([a,b]\) is Riemann integrable if and only if it is continuous almost everywhere.
1.3 Properties of Riemann Integrals
Throughout, \(f\) and \(g\) are Riemann integrable on \([a,b]\). The proofs below use the definition of the integral as a limit of Riemann sums, together with the criterion above for integrability (continuity almost everywhere).
For constants \(\alpha,\beta\) and \(a<c<b\):
- Linearity: \(\alpha f+\beta g\) is integrable and \[ \int_a^b (\alpha f(x)+\beta g(x))~dx = \alpha\int_a^bf(x)~dx + \beta \int_a^b g(x) ~ dx. \]
- Additivity over intervals: \(\displaystyle\int_a^bf=\int_a^cf+\int_c^bf\).
- Products and absolute values: \(fg\) and \(|f|\) are integrable.
Proof.
A Riemann sum is linear in the function, \(\sum(\alpha f+\beta g)(\xi_i)\Delta x_i=\alpha\sum f(\xi_i)\Delta x_i+\beta\sum g(\xi_i)\Delta x_i\), and limits are linear.
Take partitions of \([a,b]\) that contain \(c\) as a node (refining a partition by adding a point does not change the limit as the mesh tends to \(0\)). Such a Riemann sum splits into a sum over \([a,c]\) and a sum over \([c,b]\), and the limit of each piece is the corresponding integral.
If \(f\) and \(g\) are bounded and continuous almost everywhere then so is \(fg\), and \(|f|\) is continuous wherever \(f\) is, so both are integrable by the criterion.
- Monotonicity: if \(f\le g\) on \([a,b]\) then \(\int_a^bf\le\int_a^bg\). In particular \(f\ge0\) implies \(\int_a^bf\ge0\).
- Bounds: if \(m\le f\le M\) then \[ m(b-a)\le\int_a^bf\le M(b-a). \]
- Triangle inequality: \(\left|\displaystyle\int_a^bf\right|\le\displaystyle\int_a^b|f|\).
Proof.
If \(f\le g\) then every Riemann sum of \(f\) is at most the corresponding sum of \(g\), since \(\Delta x_i>0\), and the inequality is preserved in the limit.
Apply (1) to the constants \(m\le f\le M\) and use \(\int_a^bc=c(b-a)\), which is clear since every Riemann sum of a constant equals \(c(b-a)\).
We have \(-|f|\le f\le|f|\), and \(|f|\) is integrable, so by (1) and linearity \(-\int|f|\le\int f\le\int|f|\).
Orientation convention. We define \(\int_b^af:=-\int_a^bf\) for \(a<b\) and \(\int_a^af:=0\). With this, additivity \(\int_a^bf=\int_a^cf+\int_c^bf\) holds for any ordering of \(a,b,c\).
If \(f\) is continuous on \([a,b]\) there exists \(c\in[a,b]\) with \[ f(c)=\frac1{b-a}\int_a^bf(x)~dx, \] that is, \(f\) attains its average value.
Proof. Let \(m,M\) be the minimum and maximum of \(f\) on \([a,b]\), which exist since \(f\) is continuous on a compact interval. By the bounds above the average \(\frac1{b-a}\int_a^bf\) lies in \([m,M]\), and by the intermediate value theorem \(f\) takes every value in \([m,M]\).
1.3.1 Which Functions are Integrable?
By the criterion above, the following are Riemann integrable on \([a,b]\):
- every continuous function,
- every monotone function (it has at most countably many discontinuities),
- every bounded function with finitely many (or countably many) discontinuities.
Changing the values of an integrable function at finitely many points changes neither its integrability nor the value of its integral.
1.3.2 The Fundamental Theorem of Calculus
The Riemann integral and the derivative are inverse operations.
Let \(f\) be integrable on \([a,b]\).
- Define \(F(x)=\int_a^xf(t)~dt\). Then \(F\) is continuous on \([a,b]\), and if \(f\) is continuous at \(x\) then \(F\) is differentiable at \(x\) with \(F'(x)=f(x)\).
- If \(G\) is differentiable on \([a,b]\) with \(G'=f\), then \[ \int_a^bf(x)~dx=G(b)-G(a). \]
Proof.
For \(f\) bounded by \(K\), \(|F(x+h)-F(x)|=\left|\int_x^{x+h}f\right|\le K|h|\), so \(F\) is continuous. If \(f\) is continuous at \(x\) then \[ \left|\frac{F(x+h)-F(x)}{h}-f(x)\right| =\left|\frac1h\int_x^{x+h}\big(f(t)-f(x)\big)dt\right| \le\sup_{|t-x|\le|h|}|f(t)-f(x)|\to0 \] as \(h\to0\).
For any partition \(\{x_0<\dots<x_n\}\) the mean value theorem for \(G\) gives points \(\xi_i\in(x_{i-1},x_i)\) with \(G(x_i)-G(x_{i-1})=f(\xi_i)(x_i-x_{i-1})\). Summing (the left side telescopes), \[ G(b)-G(a)=\sum_{i=1}^nf(\xi_i)(x_i-x_{i-1}), \] which is a Riemann sum of \(f\). Letting \(\|\mathcal{P}\|\to0\) the right side tends to \(\int_a^bf\).
Part (2) turns the evaluation of integrals into finding antiderivatives; see Calculus of Simple Functions for tables of standard ones, and Leibniz Integral Rule for differentiating integrals that depend on a parameter.
1.4 Examples
Compute \(\int_0^1x^2~dx\) directly from Riemann sums. Since \(f(x)=x^2\) is continuous it is integrable, so the limit can be computed along any sequence of partitions with mesh \(\to0\). Take the uniform partition \(x_i=i/n\), so \(\Delta x_i=1/n\), with right endpoints \(\xi_i=i/n\): \[ S_n=\sum_{i=1}^n\left(\frac in\right)^2\frac1n=\frac1{n^3}\sum_{i=1}^ni^2 =\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6n^2}. \] As \(n\to\infty\) (mesh \(=1/n\to0\)) \[ S_n=\frac{2n^2+3n+1}{6n^2}\to\frac26=\frac13. \] This agrees with the fundamental theorem: \(\int_0^1x^2dx=\left[\tfrac{x^3}3\right]_0^1=\tfrac13\).
Let \(f=1\) on \([0,\tfrac12]\) and \(f=2\) on \((\tfrac12,1]\). It is bounded with one discontinuity, so it is integrable, and by additivity \[ \int_0^1f=\int_0^{1/2}1~dx+\int_{1/2}^12~dx=\tfrac12+2\cdot\tfrac12=\tfrac32. \] Note that \(f\) has no antiderivative on \([0,1]\) (derivatives satisfy the intermediate value property and \(f\) skips the values between \(1\) and \(2\)), yet it is integrable. So the integral exists in cases where part (2) of the fundamental theorem does not apply.
The Dirichlet function on \([0,1]\), \[ f(x)=\begin{cases}1&x\in\mathbb{Q},\\0&x\notin\mathbb{Q},\end{cases} \] is bounded but is discontinuous at every point, so it is not Riemann integrable. Directly: every subinterval of any partition contains both rational and irrational points. Choosing all sample points \(\xi_i\) rational gives every Riemann sum equal to \(\sum\Delta x_i=1\), while choosing them all irrational gives \(0\). The sums therefore do not converge to a single number as the mesh tends to \(0\).
The Lebesgue Integral handles this function: it equals \(0\) except on the countable set \(\mathbb{Q}\) of measure zero, so its Lebesgue integral is \(0\).