The Riemann Integral

Riemann integration is a topic you are likely familiar with from calculus class. Here, we are going to define the Riemann integral in a way you may not be used to. This alternate definition will allow us to explore in more depth what it means for a function to be Riemann integrable and, later, the shortcomings and alternatives to Riemann integration.

We will begin with some definitions.

Definition 1. Let $a, b \in \mathbb{R}$ with $a < b$ . A partition of $[a,b]$ is a finite list of the form $P=\{x_o,x_1,\dots ,x_n\}$ where $a=x_0<x_1<\dots < x_n = b$

This allows us to divide the interval $[a,b]$ into $n$ non-overlapping subintervals: $[a,b] = \bigcup_{k=1}^n [x_{k-1},x_k]$

Definition 2. Let $S$ be a set of real numbers. $S$ is called bounded below if there exists a real number $m \in \mathbb{R}$ , called a lower bound of S, such that $x \geq m$ for every element $x \in S$ .

$S$ is called bounded above if there exists a real number $M \in \mathbb{R}$ , called an upper bound of A, such that $x \leq M$ for every element $x \in S$ .

Definition 3. Let $S$ be a set of real numbers. The infimum or greatest lower bound of $S$ , denoted inf $\boldsymbol{S}$ , is the largest real number $t$ such that $x \geq t$ for all $x \in S$ .

The supremum or least upper bound of $S$ , denoted sup $\boldsymbol{S}$ , is the largest real number $r$ such that $x \leq r$ for all $x \in S$ .

Definition 4. Let $f$ be a real-valued function and let $A$ be some subset of the domain of $f$ . Then we define $\inf_a f = \inf \{f(x) : x \in A\}$ and $\sup_a f = \sup \{f(x) : x \in A\}$

Now, we can move on to defining the upper and lower Riemann sums.

Definition 5. Suppose $f: [a,b] \rightarrow \mathbb{R}$ is a bounded function and $P = \{ x_0, x_1, \dots , x_n\}$ is a partition of $[a,b]$ .

The lower Riemann sum is defined as the trivariate function $L(f,P,[a,b]) = \sum_{j=1}^n (x_j - x_{j-1} ) \inf_{[x_{j-1}, x_j]} f$ and the upper Riemann sum is defined as $U(f,P,[a,b]) = \sum_{j=1}^n (x_j - x_{j-1} ) \sup_{[x_{j-1}, x_j]} f$

Example 6. Define $f: [0,1] \rightarrow \mathbf{R}$ as $f(x) = x^2$ and let $P_n= \left\{ 0, \frac{1}{n}, \frac{2}{n}, \dots 1 \right\}$ .

Below are visualizations with partition $P_4$ .

For $P_n$ we have $x_j - x_{j-1} = \frac{1}{n} \quad \forall j = 1, 2, \dots, n$ Thus we have $L(x^2,P_n,[0,1]) = \sum_{j=1}^n (x_j - x_{j-1} ) \inf_{[x_{j-1}, x_j]} x^2$ $=\sum_{j=1}^n (x_j - x_{j-1} ) \left( \frac{j-1}{n}\right)^2 = \frac{1}{n} \sum_{j=1}^n \frac{(j-1)^2}{n^2} = \frac{1}{n} \sum_{j=1}^{n-1} \frac{j^2}{n^2}$ $=\frac{1}{n^3} \sum_{j=1}^{n-1} j^2$ Using the fact that $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ this summation is equal to $\frac{1}{n^2} \left(\frac{(n-1)n(2n-1)}{6}\right) = \frac{2n^2 - 3n +1}{6n^2} = L(x^2,P_n,[0,1])$

Similarly, for the upper Riemann sum: $U(x^2,P_n,[0,1]) = \frac{1}{n} \sum_{j=1}^n \frac{j^2}{n^2}$ $= \frac{2n^2 + 3n +1}{6n^2}$

Inequalities with Riemann Sums

Theorem 7. Suppose $f: [a, b] \rightarrow \mathbb{R}$ is a bounded function and $P$ , $P'$ are partitions of $[a,b]$ such that $P \in P'$ . Then $L(f, P, [a, b]) \leq L (f, P', [a, b]) \leq U(f, P, [a, b]) \leq U (f,P', [a,b])$

Proof: Suppose $P$ and $P'$ are two partitions of $[a,b]$ where $P= \{x_0,x_1,...,x_n\}$ and $P'= \{x'_0, x'_1,..., x'_N\}$ .

For each $j= 1,...,n$ there exists $K=0,..., N-1$ and $m \in \mathbb{N}$ such that $x_{j-1} = x'_K < x'_{K+1} < ... <x'_{K+m} = x_j$ . Using this, we can say that

$(x_j-x_{j-1}) \inf_{[x_{j-1},x_j]} f = \sum_{i=1}^{m} (x'_{k+i}-x'_{k+i-1}) \inf_{[x_{j-1},x_j]} f$

$\leq \sum_{i=1}^{m} (x'_{k+i}-x'_{k+i-1}) \inf_{[x'_{k+i-1},x'_{k+i}]} f$ This gives us the result that $L(f, P, [a, b]) \leq L (f, P', [a, b])$

The same argument gives us $U(f, P, [a, b]) \leq U (f,P', [a,b])$ $\blacksquare$

Theorem 8. Suppose $f: \rightarrow \mathbb{R}$ is a bounded function and $P, P'$ are any partitions of $[a,b]$ . We will always have that $L(f,P,[a,b]) \leq U(f,P', [a,b])$

The Riemann Integral

Using our new definitions of the upper and lower Riemann sums, we reach our definition of the Riemann integral.

Definition 9. Suppose $f: \rightarrow \mathbb{R}$ is a bounded function.

We define the lower Riemann integral as $L(f, [a,b]) = \sup_P L(f,P,[a,b])$ and the upper Riemann integral as $U(f, [a,b]) = \inf_P U(f,P,[a,b])$

Theorem 10. Let $f: \rightarrow \mathbb{R}$ is a bounded function. Then, it is true that $L(f, [a,b]) \leq U(f,[a,b]$

Definition 11. A bounded function on a closed interval is called Riemann integrable if $L(f, [a,b]) = U(f,[a,b]$

If $f: \rightarrow \mathbb{R}$ is Riemann integrable, then the Riemann integral is defined by $\int_a^b f = L(f, [a,b]) = U(f,[a,b]$

Example 12. Define $f: [0,1] \rightarrow \mathbf{R}$ as $f(x) = x^2$ . Show that this function is Riemann integrable.

Solution: Previously, we found that for $P_n= \left\{ 0, \frac{1}{n}, \frac{2}{n}, \dots 1 \right\}$ , the lower and upper Riemann sums are $L(x^2,P_n,[0,1]) = \frac{2n^2 - 3n +1}{6n^2}$ and $U(x^2,P_n,[0,1]) = \frac{2n^2 + 3n +1}{6n^2}$ Hence, the lower Riemann integral will be $L(x^2,[0,1]) = \sup_{n \in \mathbb{Z}^+} \frac{2n^2 - 3n +1}{6n^2}$ To find the supremum, note that this function is strictly increasing for $n \geq 1$ . Hence, the largest value $L(x^2,P_n,[0,1])$ can be is $\lim_{n \rightarrow \infty} \left( \frac{2n^2 - 3n +1}{6n^2} \right) = \frac{1}{3}$ Similarly, the upper Riemann integral is $U(x^2,[0,1]) = \inf_{n \in \mathbb{Z}^+} \frac{2n^2 + 3n +1}{6n^2}$ which is a decreasing function for $n \geq 1$ . making the infimum $\lim_{n \rightarrow \infty} \left( \frac{2n^2 + 3n +1}{6n^2} \right) = \frac{1}{3}$ Therefore, the function $f(x)=x^2$ on $[0,1]$ is Riemann integrable, and furthermore $\int_0^1 x^2 = \frac{1}{3}$

Theorem 13. Every continuous real-valued function on a closed, bounded interval is Riemann integrable.

Proof: Suppose $a,b \in \mathbb{R}$ with $a<b$ and let $f: [a, b] \to \mathbb{R}$ be a continuous function.

We are told that $f$ is continuous, but since this is true for every point in the domain, we can go further and say that $f$ is uniformly continuous. Thus we can use the definition of uniform continuity in our proof:

For every $\epsilon > 0$ , there exists $\delta > 0$ , such that if $s,t \in [a,b]$ and $|s-t| < \delta$ , then $|f(s) - f(t)| < \epsilon$ .

Now, let $n \in \mathbb{Z}^+$ be a number such that $\frac{b-a}{n} < \delta$ and let $P_n$ be the uniform (equally spaced) partition of $[a,b]$ : $a = x_0 < x_1 < x_2 < \dots < x_n = b$ with $x_b - x_{b-1} = \frac{b-a}{n}, \quad i=1, \dots, n$ We can use the facts $L(f,[a,b]) \geq L(f,P_n,[a,b])$ and $U(f,[a,b]) \leq U(f,P_n,[a,b])$ to say that $U(f,[a,b]) - L(f,[a,b]) \leq U(f,P_n,[a,b]) - L(f,P_n,[a,b])$ $= \sum_{j=1}^n \left[\left( \frac{b-a}{n}\right) \sup_{[x_{j-1},x_j]} f \right] - \sum_{j=1}^n \left[ \left( \frac{b-a}{n} \right)\inf_{[x_{j-1},x_j]} f \right]$ $= \frac{b-a}{n} \sum_{j=1}^n \left( \sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f \right)$

Note that the length of the interval $[x_{j-1}, x_j]$ is $\frac{b-a}{n}$ , and $\sup_{[x_{j-1},x_j]} f$ and $\inf_{[x_{j-1},x_j]} f$ are both values of $f(x)$ where $x_{j-1} \leq x \leq x_j$ .

Therefore, we can use uniform continuity to say that $\frac{b-a}{n} < \delta \Rightarrow \frac{\sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f}{n} < \epsilon$ So we have that $\sum_{j=1}^n \left( \sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f \right) < (b-a)\epsilon$ Since $\epsilon$ can be any infinitesimally small number, this is as good as saying $U(f,P_n,[a,b]) - L(f,P_n,[a,b]) = 0$ which means that we also have that $U(f,[a,b]) - L(f,[a,b]) = 0$ Therefore, $f$ is Riemann integrable. $\blacksquare$

Theorem 14. Suppose $f:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable. Then $(b-a) \inf_{[a,b]}f \leq \int_a^b f \leq (b-a) \sup_{[a,b]}f$

Exercises

In this section, we will use the definitions from this section to prove well-known facts about integrals. You will already know many of these to be true. However, it is beneficial to look at them through the lens of our new definition.

Problem 15. Suppose $f:[a,b] \rightarrow \mathbb{R}$ is a bounded function such that $L(f,P,[a,b]) = U(f,P,[a,b])$ for any partition $P$ of $[a,b]$ .

Prove that $f$ must be a constant function on $[a,b]$ .

Solution: Let $P= \{x_0,x_1, \dots, x_n\}$ where $a= x_0$ and $b=x_n$ . Then, since $L(f,P,[a,b]) = U(f,P,[a,b])$ we have that $\sum_{j=1}^n (x_j - x_{j-1} ) \inf_{[x_{j-1}, x_j]} f = \sum_{j=1}^n (x_j - x_{j-1} ) \sup_{[x_{j-1}, x_j]} f$ Note that this implies that $\inf_{[x_{j-1}, x_j]} f = \sup_{[x_{j-1}, x_j]} f$ Now, since this must be true independent of how we partition $[a,b]$ , it must be true that $\inf_{[a,b]} f = \sup_{[a,b]} f$ The only way that the infimum and supremum can be equal over an interval is if $f$ is constant over that interval. Therefore, $f$ must be a constant function on $[a,b]$ . $\blacksquare$

Problem 16. Let $a \leq s < t \leq b$ . Define $f:[a,b] \rightarrow \mathbb{R}$ by $f(x) = \begin{cases} 1 & \text{if } s < x < t \\ 0 & \text{otherwise} \end{cases}$

Prove that $f$ is Riemann integrable on $[a,b]$ and $\int_a^b f = t-s$

Solution: Define a partition of $[a,b]$ as $P_5 = \{a,s,s+\delta,t-\delta,t,b\}$ , where $\delta$ is some very small number. This is represented by the graph below

This gives us a lower Riemann sum of

$L(f,P_5,[a,b]) = 0(s-a) + 0(s+\delta -s) + 1(t-\delta - (s+\delta))$ $+0(t-(t-\delta)) + 0(b-t)$ $= t-s-2\delta$ and the upper Riemann sum $U(f,P_5,[a,b]) = 0(s-a) + 1(s+\delta -s) + 1(t-\delta - (s+\delta))$ $+1(t-(t-\delta)) + 0(b-t)$ $= t-s$

But are these equal? Note that $f$ is continuous for $s < x < t$ . This means that for all $\epsilon > 0$ there exists $\delta > 0$ which we can set as $\delta = \frac{\epsilon}{2}$ such that $U(f,P_5,[a,b]) - L(f,P_5,[a,b]) = 2\delta = \epsilon \quad \forall \epsilon > 0$ Therefore, these two Riemann sums are equal and $f$ is Riemann integrable. Therefore it must be true that $\int_a^b f = U(f,P_5,[a,b]) = t-s$ $\blacksquare$

Notation: We define $\mathcal{R}([a,b])$ to be the set of all Riemann integrable functions on $[a,b]$ .

Problem 17. Suppose $f:[a,b] \rightarrow \mathbb{R}$ is a bounded function. Prove that $f$ is Riemann integrable if and only if for all $\epsilon > 0$ , there exists a partition $P$ of $[a,b]$ such that $U(f,P,[a,b]) - L(f,P,[a,b]) < \epsilon$

Solution:

( $\Rightarrow$ ) Let $f \in \mathcal{R}([a,b])$ and let $\epsilon > 0$ be given. Define partitions $P_1$ and $P_2$ such that $U(f,P_2,[a,b]) - \int_a^b f < \frac{\epsilon}{2}$ and $\int_a^b f - L(f,P_1,[a,b]) < \frac{\epsilon}{2}$ Now let $P = P_1 \cup P_2$ . So we have $U(f,P,[a,b]) \leq U(f,P_2,[a,b]) < \int_a^b f + \frac{\epsilon}{2}$ $L(f,P_1,[a,b]) + \epsilon \leq L(f,P,[a,b])$ This tells us that $U(f,P,[a,b]) - L(f,P,[a,b]) < \epsilon$ ( $\Leftarrow$ ) Now let $U(f,P,[a,b]) - L(f,P,[a,b]) < \epsilon$ be given. We know, given an earlier theorem, that $L(f,P,[a,b]) \leq L(f,[a,b]) \leq U(f,[a,b]) \leq U(f,P,[a,b])$ meaning that we must also have $0 \leq U(f,[a,b]) - L(f,[a,b]) < \epsilon \quad \forall \epsilon > 0$ $\Rightarrow U(f,[a,b]) = L(f,[a,b]) = \int_a^b f$ Therefore, $f$ is Riemann integrable. $\blacksquare$

We will use this theorem to help us with many of the problems below.

Problem 18. Suppose $f,g:[a,b] \rightarrow \mathbb{R}$ are two Riemann integrable functions. Prove that $f+g$ is also Riemann integrable and that $\int_a^b (f+g) = \int_a^b f + \int_a^b g$

Solution: Let $h = f+g$ and let $P$ be a partition of $[a,b]$ .

Note that: $\inf \{f(z) | z \in [x_{j-1},x_j] \} + \inf \{g(z) | z \in [x_{j-1},x_j] \} \leq f(x) + g(x) = h$ $\Rightarrow \inf_{[x_{j-1},x_j]} f + \inf_{[x_{j-1},x_j]} g \leq \inf_{[x_{j-1},x_j]} h$ Similarly, we can deduce that $\sup_{[x_{j-1},x_j]} f + \sup_{[x_{j-1},x_j]} g \geq \sup_{[x_{j-1},x_j]} h$ Together, these two facts give us the inequalities $L(f,P,[a,b]) + L(g,P,[a,b]) \leq L(h,P,[a,b])$ $\leq U(h,P,[a,b] \leq U(f,P,[a,b] + U(g,P,[a,b]$ Since $f$ and $g$ are Riemann integrable, for all $\epsilon > 0$ , there exist partitions $P_1$ and $P_2$ such that $U(f,P_1, [a,b]) - L(f, P_1, [a,b]) < \epsilon$ and $U(g,P_2, [a,b]) - L(g, P_2, [a,b]) < \epsilon$ Now let $P = P_1 \cup P_2$ . Then we can combine the above two statements into $U(h,P_2, [a,b]) - L(h, P_2, [a,b]) < 2 \epsilon$

Therefore, $h = f+g$ is Riemann integrable.

To show that $\int_a^b (f+g) = \int_a^b f + \int_a^b g$ note that we have $\int_a^b h \leq U(h,P,[a,b]) \leq \int_a^b f + \int_a^b g + 2\epsilon$ and $\int_a^b h \geq L(h,P,[a,b]) \geq \int_a^b f + \int_a^b g + 2\epsilon n$ which tells us that $\int_a^b (f+g) = \int_a^b f + \int_a^b g$ $\blacksquare$

Problem 19. Let $f: [a,b] \rightarrow \mathbb{R}$ be Riemann integrable. Prove that $-f$ is also Riemann integrable and that $\int_a^b (-f) = - \int_a^b f$

Solution: Since $f$ is Riemann integrable, we know that $L(f,P,[a,b]) = U(f,P,[a,b])$ which clearly means that $-L(f,P,[a,b]) = -U(f,P,[a,b])$ Now, note that $\inf_{[s,t]} f = \sup_{[s,t]} (-f)$ and $\inf_{[s,t]} (-f) = \sup_{[s,t]} f$ for any interval $[s,t]$ in the domain of $f$ . (The smallest value of $f$ will be the greatest value of -f and vice versa)

Hence, we have that $-L(f,P,[a,b]) = U(-f,P,[a,b])$ and $L(-f,P,[a,b]) =- U(f,P,[a,b])$ which implies that $L(-f,P,[a,b]) = U(-f,P,[a,b])$ Therefore, $-f$ is Riemann integrable and $\int_a^b (-f) = - \int_a^b f$ $\blacksquare$

Note that we only need to adjust this proof slightly to prove that for any $c\in \mathbb{R}$ , we have $\int_a^b c f = c\int_a^b f$ We will use this fact moving forward.

Problem 20. Suppose $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable. Now, suppose that $g: [a,b] \rightarrow \mathbb{R}$ is a function such that $g(x)=f(x)$ for all except finitely many $x \in [a,b]$ . Prove that $g$ is Riemann integrable on $[a,b]$ and $\int_a^b g = \int_a^b f$

Solution: To solve this problem, first let’s look at a function with a domain of only finitely many points, and what the Riemann integral of that function would be.

Define the $d_s$ function on $[a,b]$ , with $s \in [a,b]$ as

$d_s = \begin{cases} 1 & x=s \\ 0 & x \neq s \end{cases}$

We want to show that $d_s$ is Riemann integrable and $\int_a^b d_s = 0$

Starting with the lower Riemann sum, the infimum of $d_s$ over any subinterval of $[a,b]$ is zero, so $L(d_s,P,[a,b])=0$ for any partition of $[a,b]$ and therefore $L(d_s,[a,b])=0$ Now, we look at the upper Riemann sum. Given any partition $P=\{x_0,x_1,...x_n\}$ , the supremum over each subinterval will be $\sup_{[x_{j-1,x_j}]}d_s = \begin{cases} 1 & s \in [x_{j-1},x_j] \\ 0 & \text{otherwise} \end{cases}$ and so $U(d_s,P,[a,b]) = 1 (x_{j}-x_{j-1})$ To find $U(d_s,[a,b])$ , let the length of the interval $(x_{j}-x_{j-1})$ approach zero:

$U(d_s,[a,b]) = \lim_{(x_{j}-x_{j-1}) \rightarrow 0}U(d_s,P,[a,b]) = 0$ Therefore $d_s$ is Riemann integrable and $\int_a^b d_s = 0$ Now let’s expand this function to include any finite number of nonzero points which aren’t necessarily equal to 1.

Define a set of points $\{s_1,s_2, \dots, s_k\}, \quad a \leq s_1 < s_2 < \dots < s_k \leq b$ and let $c_1, c_2, \dots, c_k$ be real numbers. Using the result from previous problems, we know that $\displaystyle \sum_{i=1}^k c_i d_{s_i}$ is Riemann integrable and $\int_a^b \sum_{i=1}^k c_i d_{s_i} = \sum_{i=1}^k c_i \int_a^b d_{s_i} = 0$ Now we can return to $g(x)$ . Let $\{s_1,s_2, \dots, s_k\}$ be the finite list of points in $[a,b]$ where $g(x) \neq f(x)$ . If we let $c_i = g(s_i) - f(s_i)$ for $i = 1,2, \dots, k$ , we can write $g(x) = f(x) + \sum_{i=1}^k c_i d_{s_i}$ $f$ and $\displaystyle \sum_{i=1}^k c_i d_{s_i}$ are integrable, so $g$ is also integrable and $\int_a^b g = \int_a^b f + \int_a^b \sum_{i=1}^k c_i d_{s_i} = \int_a^b f$ $\blacksquare$

Problem 21. Suppose $f[a,b] \rightarrow \mathbb{R}$ is a bounded function. For any $n \in \mathbb{Z}^+$ , let $P_n$ denote the partition that divides $[a,b]$ into $2^n$ intervals of equal size. Prove that

$L(f,[a,b]) = \lim_{n\rightarrow \infty} L(f,P_n,[a,b])$

and

$U(f,[a,b]) = \lim_{n\rightarrow \infty} U(f,P_n,[a,b])$

Solution: Let $P_n = \{x_0, x_1, \dots x_{2^n}\}$ where the length of each subinterval is $\Delta x_n = \frac{b-a}{2^n}$ . Let $L= \sup_{[a,b]}f$ . We will focus on the lower sum and the result for the upper sum will come as a consequence.

First, note that by definition we will have $L(f,[a,b]) \geq \lim_{n\rightarrow \infty }L(f,P_n, [a,b])$ Since $L(f,[a,b])$ is the supremum of all lower sums. So we just need to show $\Rightarrow \lim_{n\rightarrow \infty }L(f,P_n, [a,b]) \geq L(f,[a,b])$ For any $\epsilon > 0$ and partition $P = \{z_1,z_2,\dots, z_k \}$ of $[a,b]$ , choose an $n \in \mathbb{Z}^+$ such that $\left(\frac{b-a}{2^n}\right)2kL < \epsilon$ First note that if we take the partition $P\cup P_n$ , we will have $L(f,P\cup P_n, [a,b]) \geq L(f,P,[a,b])$ Now we can take the difference $|L(f,P_n,[a,b]) - L(f,P \cup P_n, [a,b])|$ $\leq \sum_{j=1}^k \left[ 2L \left( \frac{b-a}{2^n} \right) \right] = 2kL \left( \frac{b-a}{2^n} \right) < \epsilon$ This gives us the result $L(f,P_n, [a,b]) \geq L(f,P\cup P_n,[a,b]) + \epsilon$ $\Rightarrow L(f,P_n, [a,b]) > L(f,P,[a,b]) - \epsilon$ $\Rightarrow \lim_{n\rightarrow \infty }L(f,P_n, [a,b]) \geq L(f,P,[a,b]) - \epsilon$ $\Rightarrow \lim_{n\rightarrow \infty }L(f,P_n, [a,b]) \geq L(f,[a,b])$

Therefore we have that $L(f,[a,b]) = \lim_{n\rightarrow \infty }L(f,P_n, [a,b])$ We can prove the second part using the same logic. $\blacksquare$

Problem 22. Suppose $f[a,b] \rightarrow \mathbb{R}$ is Riemann integrable on $[a,b]$ . Prove that $\int_a^b f = \lim_{n\rightarrow \infty} \left[\frac{b-a}{n} \, \sum_{j=1}^n f\left(a + j \left( \frac{b-a}{n} \right)\right) \right]$

Solution: (Easier than it looks) Let $S_n = \frac{b-a}{n} \, \sum_{j=1}^n f\left(a + j \left( \frac{b-a}{n} \right)\right)$ and define a partition $P$ of $[a,b]$ as $P=\{x_0,x_1, \dots x_n\}$ where $x_j = a + j \left( \frac{b-a}{n} \right)$ for all $j = 1,2, \dots, n$ .

Then, $L(f,P_n, [a,b]) = \sum_{j=1}^n a + j \left( \frac{b-a}{n} \right) - \left[a + (j-1) \left( \frac{b-a}{n} \right)\right] \inf_{[x_{j-1},x_j]}f$ $= \frac{b-a}{n} \sum_{j=1}^n \inf_{[x_{j-1},x_j]}f \leq S_n$ Similarly, we can deduce that $S_n \geq U(f,P_n,[a,b])$ Now, look at what happens as $n\rightarrow \infty$ . We can see that $\lim_{n \rightarrow \infty} L(f,P_n, [a,b]) = \sup_P L(f,P_n, [a,b]) = L(f,[a,b]) = \int_a^b f$ and $\lim_{n \rightarrow \infty} U(f,P_n, [a,b]) = \inf_P U(f,P_n, [a,b]) = U(f,[a,b]) =$ which gives us the result $\lim_{n\rightarrow \infty} S_n = \int_a^b f$ $\blacksquare$

Problem 23. Suppose $f:[a,b]\rightarrow \mathbb{R}$ is Riemann integrable. Prove that if $c,d \in \mathbb{R}$ and $a \leq c < d \leq b$ , then f is Riemann integrable on $[c,d]$ .

Solution: Since $f$ is Riemann integrable, for all $\epsilon > 0$ , there exists a partition $P$ of $[a,b]$ such that $U(f,P,[a,b]) - L(f,P,[a,b]) < \epsilon$ Now define a new partition $P' = P \cup \{c,d \}$ since our partition of $[c,d]$ will have to contain $c$ and $d$ . Adding extra points to our partition won’t change that $U(f,P',[a,b]) - L(f,P',[a,b]) < \epsilon$ Finally, define a partition $P''$ of $[c,d]$ where $P'' = P' \cap [c,d]$ This gives us the result $U(f,P'',[c,d]) - L(f,P'',[c,d]) \leq U(f,P',[a,b]) - L(f,P',[a,b]) < \epsilon$ Hence, $f$ is integrable over $[c,d]$ . $\blacksquare$

Problem 24. Suppose $f[a,b] \rightarrow \mathbb{R}$ is a bounded function and $c \in (a,b)$ . Prove that $f$ is integrable on $[a,b]$ if and only if $f$ is integrable on $[a,c]$ and $[c,b]$ and $\int_a^bf = \int_a^c f + \int_c^b f$

Suppose $f$ is Riemann integrable on $[a,b]$ . This means that for any $\epsilon > 0$ , there exists a partition $P$ of $[a,b]$ such that $U(f,P,[a,b]) - L(f,P,[a,b]) < \epsilon$ Now, define a new partition $P' = P \cup \{c\}$ with $P'=\{x_1,x_2, \dots, x_k, \dots, x_n\}$ where $x_k = c$ and split this partition in two so that $P'_1$ is a partition of $[a,c]$ and $P'_2$ is a partition of $[c,b]$ . $L(f,P,[a,b]) = \sum_{j=1}^n (x_j-x_{j-1}) \inf_{[x-{j-1},x_j]}f$ $\sum_{j=1}^k (x_j-x_{j-1}) \inf_{[x-{j-1},x_j]}f + \sum_{j=k+1}^n (x_j-x_{j-1}) \inf_{[x-{j-1},x_j]}f$ $= L(f,P'_1,[a,c]) + L(f,P'_2,[c,b])$ Similarly, for the upper Riemann sum, we have $U(f,P,[a,b]) = U(f,P'_1,[a,c]) + U(f,P'_2,[c,b])$ Hence $U(f,P',[a,b]) - L(f,P',[a,b]) < \epsilon$ $\Rightarrow U(f,P'_1,[a,c]) + U(f,P'_2,[c,b]) - \left( L(f,P'_1,[a,c]) + L(f,P'_2,[c,b]) \right) < \epsilon$ Breaking this apart, we get $U(f,P'_1,[a,c]) - L(f,P'_1,[a,c]) < \epsilon$ and $U(f,P'_2,[c,b]) - L(f,P'_2,[c,b]) < \epsilon$ Therefore, $f$ is integrable on $[a,c]$ and $[c,b]$ .

The proof of the reverse will follow the same logic.

Now, to show that, given $f$ is integrable on $[a,c]$ , $[c,b]$ , and $[a,b]$ , $\int_a^bf = \int_a^c f + \int_c^b f$ We know that $\int_a^bf = \sup_P L(f,P,[a,b]) = \inf_P U(f,P,[a,b])$ Let’s look at the lower sum: $\sup_P L(f,P,[a,b]) = \sup_P \Big[ L(f,P_1,[a,c]) + L(f,P_2,[c,b])\Big]$ $= \sup_{P} L(f,P_1,[a,c]) + \sup_{P} L(f,P_2,[c,b])$ $= \sup_{P_1} L(f,P_1,[a,c]) + \sup_{P_2} L(f,P_2,[c,b])$ $=\int_a^c f + \int_c^b f$ $\blacksquare$

Problem 25. Suppose $f:[a,b]\rightarrow \mathbb{R}$ is Riemann integrable. Define $F(t) = \begin{cases} 0 & t=a \\ \displaystyle \int_a^t f & t \neq a \end{cases}$ Prove that $F$ is continuous.

Solution: To prove continuity, we will show that $\lim_{t \rightarrow c} F(t) = 0 \quad \forall c \in [a,b]$ We will split this into the cases $c=a$ and $c \neq a$

For $c=a$ , we need to show that $\lim_{t \rightarrow a^+} F(t) = 0$ Since $f$ is Riemann integrable on $[a,b]$ , given any $\epsilon > 0$ , there exists a $\delta >0$ such that if $0 < t-a < \delta$ , then $\left| \int_a^b f \right| < \epsilon$ This is enough to tell us that $\lim_{t \rightarrow a^+} F(t) = 0$ Now, let $c \in (a,b]$ and let’s show that $\lim_{t \rightarrow c} F(t) = \lim_{t \rightarrow c} \int_a^c f = 0$ (For $c=b$ this will be the left-sided limit only, but the argument is otherwise identical so we will omit it.) We know that since $f$ is Riemann integrable on $[a,b]$ , it must be bounded on $[a,b]$ . Let $M$ be the upper bound. Our result from Problem 1.2 gives us that $\left| \int_a^c f \right| \leq M|t-c|$ So we have that $\lim_{t \rightarrow c} \int_a^c f = \lim_{t \rightarrow c} M|t-c| = 0$ Hence, $F(t)$ is continuous. $\blacksquare$

Problem 26. Suppose $f:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable. Prove that $|f|$ is Riemann integrable and that $\left| \int_a^b f \right| \leq \int_a^b |f|$

Solution: Since $f$ is Riemann integrable, we know that, given $\epsilon > 0$ , there exists a partition $P$ such that $U(f,P, [a,b]) - L(f,P, [a,b]) < \epsilon$ Note that if we take the absolute value of $f$ , we have $\sup_{[x_{j-1},x_j} |f| - \inf_{[x_{j-1},x_j} |f| \leq \sup_{[x_{j-1},x_j} f - \inf_{[x_{j-1},x_j} f$ for any interval $[x_{j-1},x_j]$ contained in $[a,b]$ . So we have $U(|f|,P, [a,b]) - L(|f|,P, [a,b]) \leq U(f,P, [a,b]) - L(f,P, [a,b]) < \epsilon$ Therefore $|f|$ is Riemann integrable.

To prove the second part, define $c$ such that $|c| = 1$ . (In other words, $c=1$ or $c=-1$ .) We know that $\int_a^b c f = c \int_a^b f$ Taking the absolute value of both sides we get $\left| \int_a^b c f \right|= \left| c \int_a^b f \right| = |c| \left| \int_a^b f \right| = \left| \int_a^b f \right|$ Now, using $cf \leq |f|$ , we can conclude $\left| \int_a^b f \right| = \left| \int_a^b c f \right| \leq \int_a^b |f|$ $\blacksquare$

Problem 27. Suppose $f:[a,b] \rightarrow \mathbb{R}$ is an increasing function. Prove that $f$ is Riemann integrable.

Solution: Let’s choose a partition and show that the difference between the upper and lower sum goes to zero as the number of subintervals in our partition goes to infinity.

Let $P_n = \{ a + \left(\frac{b-a}{n} \right)k : k=0,1, \dots, n\}$ . Note that since $f$ is an increasing function, the infimum over any interval $[x_{j-1}, x_j]$ in $[a,b]$ will be $f(x_{j-1})$ while the supremum will be $f(x_{j})$ . So we have: $U(f,P_n,[a,b]) - L(f,P_n,[a,b])$ $= \sum_{j=1}^n (f(x_j) - f(x_{j-1})) (x_j - x_{j-1})$ $= \left( \frac{b-a}{n} \right) \sum_{j=1}^n (f(x_j) - f(x_{j-1}))$ $\left( \frac{b-a}{n} \right) (f(b) - f(a))$ Since $a$ and $b$ are fixed $\lim_{n \rightarrow \infty} \left( \frac{b-a}{n} \right) (f(b) - f(a)) = 0$ Therefore $f$ is Riemann integrable. $\blacksquare$

Problem 28. Suppose $f_1, f_2, \dots$ is a sequence of Riemann integrable functions on $[a,b]$ which converges uniformly to a function $f$ as $n \rightarrow \infty$ .

Prove that $f$ is Riemann integrable on $[a,b]$ and $\int_a^b f = \lim_{n \rightarrow \infty} \int_a^b f_n$

Solution: Uniform convergence tells us that for any $\epsilon >0$ , there exists an $N \in \mathbb{Z}^+$ such that for all $n \geq N$ $\sup_{[a,b]} |f_n(x) - f(x)| < \epsilon$

Let $\displaystyle \epsilon_n = \sup_{[a,b]} |f_n(x) - f(x)|$ . In other words, $f_n - \epsilon_n \leq f \leq f_n + \epsilon_n$ Therefore, we have that $\int_a^b (f_n - \epsilon_n) \leq L(f,[a,b]) \leq U(f,[a,b]) \leq \int_a^b (f_n + \epsilon_n)$ $\Rightarrow U(f,[a,b]) - L(f,[a,b]) \leq 2 \epsilon (b-a)$ which will go to zero as $n\rightarrow \infty$ . Therefore, $f$ is Riemann integrable.

Furthermore, if we choose $n \geq N$ we have $\int_a^b f - \int_a^b f_n = \int_a^b f - f_n \leq \int_a^b \epsilon = (b-a)\epsilon$ which gives us the result $\int_a^b f = \lim_{n \rightarrow \infty} \int_a^b f_n$ $\blacksquare$

E

H

I

L

P

R

W

Riemann Integration

Part 1

The Riemann Integral

Inequalities with Riemann Sums

The Riemann Integral

Exercises