21 Dirichlet's Function, Integrability

Lecture from: 13.05.2024 | Video: Video ETHZ

Review: The Road to Riemann Integrability

Quick Recap: Our Toolkit for Defining Area Just to refresh, here’s the conceptual path we’ve taken to define the area under a curve $f(x)$ on an interval $[a,b]$:

1. Bounded Function: We start with a function $f$ that doesn’t shoot off to infinity on $I=[a,b]$.
2. Partition: We slice the interval $I$ into smaller pieces: $P=\{a=x_0,x_1,\dots ,x_n=b\}$.
3. Local Extrema: On each tiny piece $[x_{i-1},x_i]$, we find the “floor” $m_i=\inf f(x)$ and the “ceiling” $M_i=\sup f(x)$ of the function.
4. Lower Sum: $s(f,P)=\sum m_i(x_i-x_{i-1})$ is the sum of areas of rectangles guaranteed to be under the curve—an underestimate of the total area.
5. Upper Sum: $S(f,P)=\sum M_i(x_i-x_{i-1})$ is the sum of areas of rectangles guaranteed to be over (or containing) the curve—an overestimate.
6. Refinement: Making the slices smaller (refining $P$ to $P^{\prime }$) squeezes these sums: $s(f,P)\le s(f,P^{\prime })\le S(f,P^{\prime })\le S(f,P)$. The gap narrows!
7. Fundamental Bound: Any lower sum is always less than or equal to any upper sum: $s(f,P_1)\le S(f,P_2)$.
8. Best Underestimate: The lower Darboux integral $s(f)={\sup }_{P}s(f,P)$ is the “highest possible” lower sum.
9. Best Overestimate: The upper Darboux integral $S(f)={\inf }_{P}S(f,P)$ is the “lowest possible” upper sum.
10. Integrability! If these best estimates meet ($s(f)=S(f)$), the function $f$ is Riemann integrable. This common value is the Riemann integral ${\int }_a^{b}f(x)dx$, representing the exact area.

Theorem: A Practical Criterion for Riemann Integrability

A Handy Test for Integrability Calculating the lower and upper Darboux integrals ($s(f)$ and $S(f)$) by considering all possible partitions is practically impossible. This theorem provides a much more manageable test: a bounded function $f$ is Riemann integrable if and only if we can find at least one partition $P$ for which its upper sum $S(f,P)$ and lower sum $s(f,P)$ are incredibly close to each other—specifically, their difference can be made smaller than any pre-chosen positive number $ϵ$. If you can squeeze the gap this tight for some partition, the function is integrable!

A bounded function $f:I\to \mathbb{R}$ (where $I=[a,b]$) is Riemann integrable if and only if for every $ϵ>0$, there exists a partition $P\in \mathcal{P}(I)$ such that: $S(f,P)-s(f,P)<ϵ$

Proof

• "$⟹$" (If $f$ is integrable, then for every $ϵ>0$, such a $P$ exists): Suppose $f$ is Riemann integrable. This means $s(f)=S(f)$, where $s(f)$ is the supremum of all lower sums and $S(f)$ is the infimum of all upper sums. Let $ϵ>0$ be given. By the definition of supremum, since $s(f)={\sup }_{P}s(f,P)$, there must exist some partition $P_1$ such that $s(f,P_1)$ is very close to $s(f)$, specifically $s(f,P_1)>s(f)-ϵ/2$. Similarly, by the definition of infimum, since $S(f)={\inf }_{P}S(f,P)$, there must exist some partition $P_2$ such that $S(f,P_2)<S(f)+ϵ/2$. Now, let $P=P_1\cup P_2$ be the common refinement of $P_1$ and $P_2$. We know that refining a partition does not decrease the lower sum and does not increase the upper sum (Lemma 5.1.2 (1)). So: $s(f,P)\ge s(f,P_1)>s(f)-ϵ/2$ $S(f,P)\le S(f,P_2)<S(f)+ϵ/2$ Subtracting the (negated) inequality for $s(f,P)$ from the inequality for $S(f,P)$: $S(f,P)-s(f,P)<(S(f)+ϵ/2)-(s(f)-ϵ/2)$. Since we assumed $f$ is integrable, $S(f)=s(f)$. So, the expression becomes: $S(f,P)-s(f,P)<(s(f)+ϵ/2)-(s(f)-ϵ/2)=ϵ/2+ϵ/2=ϵ$. Thus, for this common refinement $P$, the condition $S(f,P)-s(f,P)<ϵ$ holds.

• "$⟸$" (If for every $ϵ>0$, such a $P$ exists, then $f$ is integrable): Suppose that for every $ϵ>0$, there exists a partition $P\in \mathcal{P}(I)$ such that $S(f,P)-s(f,P)<ϵ$. We always know that $s(f,P)\le s(f)$ (by definition of $s(f)$ as supremum) and $S(f)\le S(f,P)$ (by definition of $S(f)$ as infimum). Also, we know $s(f)\le S(f)$. Combining these, we have $S(f)-s(f)\le S(f,P)-s(f,P)$. Given our assumption, this means $S(f)-s(f)<ϵ$. This inequality, $0\le S(f)-s(f)<ϵ$, holds for every positive $ϵ$. The only non-negative number that is smaller than every positive number is 0. Therefore, $S(f)-s(f)=0$, which implies $S(f)=s(f)$. By definition, this means $f$ is Riemann integrable.

Example: Integrating $f(x)=x$ on $[a,b]$

Let’s Calculate an Integral from Scratch: Area Under $f(x)=x$ We’ll now apply the Darboux sum machinery to a simple function, $f(x)=x$, on an interval $[a,b]$. The goal is to see if its lower and upper sums converge to the same value as we make our partition finer and finer. We intuitively expect the area to be that of a trapezoid.

Let $f(x)=x$ on the interval $[a,b]$.

We’ll use a sequence of partitions $P_n$, where each $P_n$ divides $[a,b]$ into $n$ subintervals of equal length.

The total length of the interval is $b-a$. So, the length of each subinterval is $h=\frac{b-a}{n}$. The points of the partition are $x_i=a+i\cdot h$ for $i=0,1,\dots ,n$.

Thus, $x_0=a$, $x_1=a+h$, $x_2=a+2h$, …, $x_n=a+nh=a+n\left(\frac{b-a}{n}\right)=a+(b-a)=b$.

The $i$-th subinterval is $[x_{i-1},x_i]=[a+(i-1)h,a+ih]$. Its length is $\Delta x_i=x_i-x_{i-1}=h$.

Since $f(x)=x$ is a strictly increasing function:

• On any subinterval $[x_{i-1},x_i]$, the minimum value (infimum) is at the left endpoint: $m_i=f(x_{i-1})=x_{i-1}=a+(i-1)h$.
• On any subinterval $[x_{i-1},x_i]$, the maximum value (supremum) is at the right endpoint: $M_i=f(x_i)=x_i=a+ih$.

Lower Darboux Sum $s(f,P_n)$

Let $j=i-1$. As $i$ ranges from $1$ to $n$, $j$ ranges from $0$ to $n-1$.

Using the arithmetic series sum formula ${\sum }_{j=0}^{k}j=\frac{k(k+1)}{2}$, we have ${\sum }_{j=0}^{n-1}j=\frac{(n-1)n}{2}$.

Now, substitute $h=\frac{b-a}{n}$:

Upper Darboux Sum $S(f,P_n)$

Using the formula ${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$:

Substitute $h=\frac{b-a}{n}$:

Taking the Limit as $n\to \infty$ (mesh size goes to 0)

As $n\to \infty$, the term $\frac{1}{n}\to 0$. ${\lim }_{n\to \infty }s(f,P_n)=(b-a)a+\frac{(b-a{)}^2}{2}(1-0)=(b-a)a+\frac{(b-a{)}^2}{2}$. ${\lim }_{n\to \infty }S(f,P_n)=(b-a)a+\frac{(b-a{)}^2}{2}(1+0)=(b-a)a+\frac{(b-a{)}^2}{2}$.

Let’s simplify this common limit value: $(b-a)a+\frac{(b-a{)}^2}{2}=(b-a)\left(a+\frac{b-a}{2}\right)=(b-a)\left(\frac{2a+b-a}{2}\right)=(b-a)\frac{a+b}{2}=\frac{b^2-a^2}{2}$.

To formally show integrability using the criterion $S(f,P_n)-s(f,P_n)<ϵ$: $S(f,P_n)-s(f,P_n)=\left((b-a)a+\frac{(b-a{)}^2}{2}\left(1+\frac{1}{n}\right)\right)-\left((b-a)a+\frac{(b-a{)}^2}{2}\left(1-\frac{1}{n}\right)\right)$ $=\frac{(b-a{)}^2}{2}\left((1+\frac{1}{n})-(1-\frac{1}{n})\right)=\frac{(b-a{)}^2}{2}\left(\frac{2}{n}\right)=\frac{(b-a{)}^2}{n}$.

For any $ϵ>0$, we can choose $n$ large enough (specifically $n>\frac{(b-a{)}^2}{ϵ}$) such that $\frac{(b-a{)}^2}{n}<ϵ$.

Thus, by Theorem 5.1.4, $f(x)=x$ is integrable.

Since the limit of the lower sums equals the limit of the upper sums (and the difference $S(f,P_n)-s(f,P_n)\to 0$), it follows that $s(f)=S(f)=\frac{b^2-a^2}{2}$.

Therefore, $f(x)=x$ is Riemann integrable on $[a,b]$ and

This result perfectly matches the geometric formula for the area of a trapezoid with bases $f(a)=a$ and $f(b)=b$ and height $b-a$: Area $=\frac{1}{2}(a+b)(b-a)=\frac{b^2-a^2}{2}$.

Clicker Question: ${\int }_{-1}^1\mathrm{sgn}(x)dx=?$

Integrating a Jumpy Function: The Sign Function sgn(x) The sign function $\mathrm{sgn}(x)$ is $-1$ for $x<0$, $0$ for $x=0$, and $1$ for $x>0$. It has a jump discontinuity at $x=0$. Can we still define its integral on $[-1,1]$? Let’s investigate with a specific sequence of partitions that “squeeze” the jump.

Consider the sign function $\mathrm{sgn}(x)=\{\begin{array}{ll}-1& \text{if }x<0\\ 0& \text{if }x=0\\ 1& \text{if }x>0\end{array}$ on the interval $[-1,1]$.

Let’s use a sequence of partitions $P_n=\{-1,-1/n,1/n,1\}$ for $n>1$ (so $-1/n<1/n$).

The subintervals are $I_1=[-1,-1/n]$, $I_2=[-1/n,1/n]$, $I_3=[1/n,1]$.

Lengths: $\Delta x_1=-1/n-(-1)=1-1/n$. $\Delta x_2=1/n-(-1/n)=2/n$. $\Delta x_3=1-1/n$.

• On $I_1=[-1,-1/n]$ (for $x<0$): $f(x)=-1$. So $m_1=-1,M_1=-1$.
• On $I_2=[-1/n,1/n]$: The interval contains $x<0$ (where $f(x)=-1$), $x=0$ (where $f(x)=0$), and $x>0$ (where $f(x)=1$). So $m_2=\inf \{f(x)\mid x\in [-1/n,1/n]\}=-1$. And $M_2=\sup \{f(x)\mid x\in [-1/n,1/n]\}=1$.
• On $I_3=[1/n,1]$ (for $x>0$): $f(x)=1$. So $m_3=1,M_3=1$.

Lower Darboux Sum $s(\mathrm{sgn},P_n)$: $s(\mathrm{sgn},P_n)=m_1\Delta x_1+m_2\Delta x_2+m_3\Delta x_3$ $=(-1)(1-1/n)+(-1)(2/n)+(1)(1-1/n)$ $=-1+1/n-2/n+1-1/n=-2/n$.

Upper Darboux Sum $S(\mathrm{sgn},P_n)$: $S(\mathrm{sgn},P_n)=M_1\Delta x_1+M_2\Delta x_2+M_3\Delta x_3$ $=(-1)(1-1/n)+(1)(2/n)+(1)(1-1/n)$ $=-1+1/n+2/n+1-1/n=2/n$.

Now, we look at the difference: $S(\mathrm{sgn},P_n)-s(\mathrm{sgn},P_n)=(2/n)-(-2/n)=4/n$. For any $ϵ>0$, we can choose $n$ large enough (specifically $n>4/ϵ$) such that $4/n<ϵ$. By Theorem 5.1.4, this implies that $\mathrm{sgn}(x)$ is Riemann integrable on $[-1,1]$.

To find the value of the integral, we can take the limit of, say, the lower sums (or upper sums): ${\lim }_{n\to \infty }s(\mathrm{sgn},P_n)={\lim }_{n\to \infty }(-2/n)=0$. ${\lim }_{n\to \infty }S(\mathrm{sgn},P_n)={\lim }_{n\to \infty }(2/n)=0$. Since these limits are equal, the value of the integral is 0. ${\int }_{-1}^1\mathrm{sgn}(x)dx=0$ This makes intuitive sense: the “negative area” of $-1$ on $[-1,0)$ seems to cancel the “positive area” of $+1$ on $(0,1]$. The single point at $x=0$ does not contribute to the integral’s value for Riemann integration.

Example: Dirichlet’s Function - Not Riemann Integrable

A Function That Defies Integration: Dirichlet’s Function Consider a very “pathological” function: $f(x)=1$ if $x$ is rational, and $f(x)=0$ if $x$ is irrational. This is Dirichlet’s function. We’ll see that it’s impossible to define its area using Riemann sums because the lower and upper sums never get close.

Let $f:[0,1]\to \mathbb{R}$ be defined as $f(x)=\{\begin{array}{ll}1& \text{if }x\text{ is rational}\\ 0& \text{if }x\text{ is irrational}\end{array}$.

Consider any partition $P=\{0=x_0,x_1,\dots ,x_n=1\}$ of the interval $[0,1]$.

For any subinterval $[x_{i-1},x_i]$ (which has a positive length):

• Since rational numbers are dense in $\mathbb{R}$, this subinterval $[x_{i-1},x_i]$ must contain at least one rational number. Thus, $f(x)=1$ for some $x$ in the subinterval.
• Since irrational numbers are also dense in $\mathbb{R}$, this subinterval $[x_{i-1},x_i]$ must contain at least one irrational number. Thus, $f(x)=0$ for some $x$ in the subinterval.

Therefore, for each $i=1,\dots ,n$:

• The infimum is $m_i=\inf \{f(x)\mid x\in [x_{i-1},x_i]\}=0$ (because there are irrationals where $f(x)=0$).
• The supremum is $M_i=\sup \{f(x)\mid x\in [x_{i-1},x_i]\}=1$ (because there are rationals where $f(x)=1$).

So, for any partition $P$: The lower Darboux sum is $s(f,P)={\sum }_{i=1}^n{m}_i(x_i-x_{i-1})={\sum }_{i=1}^{n}0\cdot (x_i-x_{i-1})=0$.

The upper Darboux sum is $S(f,P)={\sum }_{i=1}^n{M}_i(x_i-x_{i-1})={\sum }_{i=1}^{n}1\cdot (x_i-x_{i-1})=(x_1-x_0)+\cdots +(x_n-x_{n-1})=x_n-x_0=1-0=1$.

This means: The lower Darboux integral is $s(f)={\sup }_{P}s(f,P)={\sup }_P\{0\}=0$. The upper Darboux integral is $S(f)={\inf }_{P}S(f,P)={\inf }_P\{1\}=1$. Since $s(f)=0$ and $S(f)=1$, we have $s(f)\ne S(f)$.

Therefore, the Dirichlet function $f$ is not Riemann integrable on $[0,1]$. The gap $S(f,P)-s(f,P)=1-0=1$ can never be made smaller than any $ϵ<1$.

Riemann’s Integrability Criterion (using Mesh Size)

A Stronger Test: Uniform Closeness with Small Mesh This criterion, also due to Riemann, provides an alternative characterization of integrability. It says a function is integrable if and only if by making the mesh of the partition (the width of the widest subinterval) sufficiently small, we can guarantee that the difference $S(f,P)-s(f,P)$ becomes arbitrarily small for any such fine partition. This is a more “uniform” condition.

We now consider what happens when we control the maximum size of the subintervals in a partition. For a partition $P=\{a=x_0,x_1,\dots ,x_n=b\}$ of $I=[a,b]$, we define its mesh (or norm, or “Maschenweite” in German) as the length of the largest subinterval: $\delta (P):={\max }_{1\le i\le n}(x_i-x_{i-1})$ Let ${\mathcal{P}}_{\delta }(I)$ denote the set of all partitions $P$ of $I$ with mesh $\delta (P)<\delta$ (meaning every subinterval in $P$ has length less than $\delta$).

Theorem: Riemann’s Criterion for Integrability

A bounded function $f:I\to \mathbb{R}$ is Riemann integrable on $I=[a,b]$ if and only if for every $ϵ>0$, there exists a $\delta >0$ such that for all partitions $P\in {\mathcal{P}}_{\delta }(I)$ (i.e., for all partitions $P$ whose mesh $\delta (P)$ is less than $\delta$): $S(f,P)-s(f,P)<ϵ$

Proof Sketch

• "$⟹$" (If $f$ is integrable, then the condition holds): This is the more involved direction. Assume $f$ is integrable. By Theorem 5.1.4, for a given ${ϵ}^{\prime }=ϵ/2$, there exists a specific partition $P_0=\{y_0,\dots ,y_m\}$ such that $S(f,P_0)-s(f,P_0)<ϵ/2$.

  Now, consider any partition $Q$ with a very small mesh, $\delta (Q)$. Let $P^{\ast }=P_0\cup Q$ be their common refinement. We know $S(f,P^{\ast })-s(f,P^{\ast })\le S(f,P_0)-s(f,P_0)<ϵ/2$.

  The challenge is to relate $S(f,Q)-s(f,Q)$ to $S(f,P^{\ast })-s(f,P^{\ast })$. The partition $P^{\ast }$ is obtained from $Q$ by adding at most $m-1$ interior points from $P_0$.

  When a subinterval of $Q$ is split by a point from $P_0$, the term $M_i-m_i$ for that original subinterval in $Q$ is replaced by two (or more) terms in $P^{\ast }$.

  The difference $(S(f,Q)-s(f,Q))-(S(f,P^{\ast })-s(f,P^{\ast }))$ can be bounded. Each of the $m-1$ interior points of $P_0$ can increase the difference $S(f,P)-s(f,P)$ by at most $2(M_{glob}-m_{glob})\delta (Q)$ when going from $P^{\ast }$ back to $Q$ (roughly, where $M_{glob},m_{glob}$ are global bounds for $f$).

  

  If we choose $\delta (Q)$ small enough, specifically $\delta (Q)<\frac{ϵ}{4m(M_{glob}-m_{glob})}$ (if $M_{glob}\ne m_{glob}$), the total contribution from these “differences” due to the $m-1$ points can be made less than $ϵ/2$.

  Then $S(f,Q)-s(f,Q)<(S(f,P^{\ast })-s(f,P^{\ast }))+ϵ/2<ϵ/2+ϵ/2=ϵ$. This makes it plausible that making the mesh $\delta (Q)$ small enough forces $S(f,Q)-s(f,Q)$ to be small.

• "$⟸$" (If the condition holds, then $f$ is integrable): This direction is more straightforward. If for every $ϵ>0$, we can find a $\delta >0$ such that any partition $P$ with mesh $\delta (P)<\delta$ satisfies $S(f,P)-s(f,P)<ϵ$, then we can certainly find one such partition $P$. This satisfies the condition of Theorem 5.1.4, which then implies that $f$ is Riemann integrable.

This theorem is powerful because it links integrability to the behavior of sums for all sufficiently fine partitions, not just the existence of one.

Corollary: Riemann Sums Converging to the Integral

The Riemann Sum: The Most Intuitive Definition This is perhaps the most direct and intuitive way to think about the integral, especially when first learning calculus. If a function $f$ is Riemann integrable, it means that if you slice the interval $[a,b]$ into small pieces (making the mesh $\delta (P)\to 0$), pick any sample point ${\xi }_i$ within each piece $[x_{i-1},x_i]$, and form the sum $\sum f({\xi }_i)\Delta x_i$, this sum will get arbitrarily close to the definite value of the integral ${\int }_a^{b}f(x)dx$. The amazing part is that it doesn’t matter which sample point ${\xi }_i$ you choose in each subinterval, as long as the subintervals are small enough.

A bounded function $f:[a,b]\to \mathbb{R}$ is Riemann integrable with integral $A={\int }_a^{b}f(x)dx$ if and only if for every $ϵ>0$, there exists a $\delta >0$ such that for every partition $P=\{a=x_0,\dots ,x_n=b\}$ with mesh $\delta (P)<\delta$, and for any choice of intermediate points (sample points) ${\xi }_i\in [x_{i-1},x_i]$ for each $i=1,\dots ,n$: $\left|A-{\sum }_{i=1}^{n}f({\xi }_i)(x_i-x_{i-1})\right|<ϵ$ The sum $R(f,P,\{{\xi }_i\})={\sum }_{i=1}^{n}f({\xi }_i)(x_i-x_{i-1})$ is called a Riemann Sum for $f$ corresponding to the partition $P$ and the choice of points $\{{\xi }_i\}$.

In simpler terms: “When the mesh size of the partition goes to zero, then the Riemann sums converge to the integral, independently of the choice of intermediate points ${\xi }_i$.”

This is because for any choice of ${\xi }_i$, $m_i\le f({\xi }_i)\le M_i$, so $s(f,P)\le \sum f({\xi }_i)\Delta x_i\le S(f,P)$. If $s(f,P)$ and $S(f,P)$ both converge to $A$, then by the Squeeze Theorem, the Riemann sum must also converge to $A$.

Integrable Functions

What kinds of functions are actually Riemann integrable?

Theorem: Properties of Integrable Functions (Algebraic Operations)

Integrability Plays Nicely with Algebra If you have functions $f$ and $g$ that are Riemann integrable on an interval $[a,b]$, then many common combinations of these functions are also integrable on that interval:

• Sums/Differences: $f+g$
• Constant Multiples: $\lambda f$
• Products: $f\cdot g$
• Absolute Values: $|f|$
• Max/Min: $\max (f,g)$ and $\min (f,g)$
• Quotients: $f/g$ (as long as $g(x)$ stays safely away from zero). Furthermore, the integral itself behaves linearly: the integral of a sum is the sum of the integrals, and constants can be pulled out.

Let $f,g:[a,b]\to \mathbb{R}$ be bounded and Riemann integrable functions, and let $\lambda \in \mathbb{R}$ be a constant. Then the following functions are also Riemann integrable on $[a,b]$:

• $f+g$ (sum)
• $\lambda f$ (scalar multiple)
• $f\cdot g$ (product)
• $|f|$ (absolute value)
• $\max (f,g)$ and $\min (f,g)$
• If there exists a $\beta >0$ such that $|g(x)|\ge \beta$ for all $x\in [a,b]$ (i.e., $g(x)$ is bounded away from zero), then the quotient $f/g$ is also Riemann integrable on $[a,b]$.

The integral operator is linear, meaning:

1. Additivity: ${\int }_a^b(f(x)+g(x))dx={\int }_a^{b}f(x)dx+{\int }_a^{b}g(x)dx$
2. Homogeneity: ${\int }_a^b(\lambda f(x))dx=\lambda {\int }_a^{b}f(x)dx$

Proof (for additivity of the integral, demonstrating integrability of $f+g$ using Riemann Sums)

We will prove that if $f$ and $g$ are integrable, then $f+g$ is integrable and its integral is the sum of the integrals of $f$ and $g$. We use the Riemann Sum convergence criterion (Corollary to Theorem 5.1.8).

Let $A_f={\int }_a^{b}f(x)dx$ and $A_g={\int }_a^{b}g(x)dx$.

Let $ϵ>0$ be given. Since $f$ is integrable, there exists a ${\delta }_1>0$ such that for any partition $P$ with mesh $\delta (P)<{\delta }_1$ and any choice of ${\xi }_i\in [x_{i-1},x_i]$: $\left|A_f-{\sum }_{i=1}^{n}f({\xi }_i)(x_i-x_{i-1})\right|<ϵ/2$.

Similarly, since $g$ is integrable, there exists a ${\delta }_2>0$ such that for any partition $P$ with mesh $\delta (P)<{\delta }_2$ and any choice of ${\xi }_i$: $\left|A_g-{\sum }_{i=1}^{n}g({\xi }_i)(x_i-x_{i-1})\right|<ϵ/2$.

Let $\delta =\min ({\delta }_1,{\delta }_2)$. Now, for any partition $P=\{x_0,\dots ,x_n\}$ with $\delta (P)<\delta$ and any choice of sample points ${\xi }_i$:

Both of the above inequalities hold. Consider the Riemann sum for $f+g$: ${\sum }_{i=1}^n(f({\xi }_i)+g({\xi }_i))(x_i-x_{i-1})={\sum }_{i=1}^{n}f({\xi }_i)(x_i-x_{i-1})+{\sum }_{i=1}^{n}g({\xi }_i)(x_i-x_{i-1})$ We want to show this sum is close to $A_f+A_g$: $\left|(A_f+A_g)-{\sum }_{i=1}^n(f({\xi }_i)+g({\xi }_i))(x_i-x_{i-1})\right|$ $=\left|(A_f-{\sum }_{i=1}^{n}f({\xi }_i)(x_i-x_{i-1}))+(A_g-{\sum }_{i=1}^{n}g({\xi }_i)(x_i-x_{i-1}))\right|$ By the triangle inequality ($|X+Y|\le |X|+|Y|$): $\le \left|A_f-{\sum }_{i=1}^{n}f({\xi }_i)(x_i-x_{i-1})\right|+\left|A_g-{\sum }_{i=1}^{n}g({\xi }_i)(x_i-x_{i-1})\right|$ $<ϵ/2+ϵ/2=ϵ$ Since for every $ϵ>0$, we found a $\delta >0$ such that the Riemann sum for $f+g$ is within $ϵ$ of $A_f+A_g$, the Corollary implies that $f+g$ is integrable and its integral is $A_f+A_g$.

(Proofs for other properties like product $f\cdot g$ are more involved, often using $S(f,P)-s(f,P)<ϵ$).

Corollary: Integrability of Polynomials and Rational Functions

Common Functions That Are Integrable Thanks to the properties above, and knowing that constant functions and $f(x)=x$ are integrable:

• All polynomials are Riemann integrable on any closed, bounded interval $[a,b]$.
• Rational functions (a polynomial divided by another polynomial) are Riemann integrable on any closed, bounded interval $[a,b]$ as long as the denominator polynomial has no roots (is never zero) within that interval.

• If $P(x)$ is a polynomial and $Q(x)$ is a polynomial such that $Q(x)\ne 0$ for all $x\in [a,b]$, then the rational function $R(x)=P(x)/Q(x)$ is integrable on $[a,b]$. (This follows because such a rational function is continuous on $[a,b]$, and we’ll soon see continuous functions are integrable. Alternatively, it builds from integrability of $x^n$ and product/quotient rules).
• In particular, every polynomial $P(x)$ (which can be seen as $P(x)/1$, where $Q(x)=1$ has no roots) is integrable on $[a,b]$.

Clicker Question: Relationship between $\int (fg)$ and $(\int f)(\int g)$

Integral of a Product vs. Product of Integrals: Any Rules? We know the integral of a sum is the sum of the integrals. Does a similar rule hold for products? Is $\int (fg)$ related to $(\int f)(\int g)$ in a simple way? Let’s test with an easy example.

Let $f,g:[a,b]\to [0,\infty )$ be bounded and integrable (so they are non-negative, just to keep it simple).

What is the relationship between ${\int }_a^{b}f(x)g(x)dx$ and $\left({\int }_a^{b}f(x)dx\right)\left({\int }_a^{b}g(x)dx\right)$?

Consider $f(x)=1$ and $g(x)=1$ on an interval $[a,b]$.

${\int }_a^{b}f(x)g(x)dx={\int }_a^b(1\cdot 1)dx={\int }_a^{b}1dx$. The integral of 1 over $[a,b]$ is the area of a rectangle with height 1 and width $(b-a)$, so it’s $(b-a)$.

Now, the product of the integrals: $\left({\int }_a^{b}f(x)dx\right)\left({\int }_a^{b}g(x)dx\right)=\left({\int }_a^{b}1dx\right)\left({\int }_a^{b}1dx\right)=(b-a)\cdot (b-a)=(b-a{)}^2$.

Is $(b-a)$ generally equal to, less than, or greater than $(b-a{)}^2$?

• If $b-a>1$ (e.g., interval $[0,2]$), then $(b-a{)}^2>(b-a)$. (e.g., $2^2=4>2$).
• If $0<b-a<1$ (e.g., interval $[0,1/2]$), then $(b-a{)}^2<(b-a)$. (e.g., $(1/2{)}^2=1/4<1/2$).
• If $b-a=1$ (e.g., interval $[0,1]$), then $(b-a{)}^2=(b-a)$. (e.g., $1^2=1$).
• If $b-a=0$ (interval is a point), both are $0$.

This simple example shows that there is no general universally valid inequality or equality that holds in all cases between $\int (fg)$ and $(\int f)(\int g)$ without further conditions (like Cauchy-Schwarz inequality for integrals, which is $(\int fg{)}^2\le (\int f^2)(\int g^2)$).

Integrability of Continuous Functions

A Very Important Class of Integrable Functions: Continuous Ones! This is a cornerstone result in Riemann integration: If a function is continuous over a closed and bounded interval $[a,b]$, then it is guaranteed to be Riemann integrable on that interval. This means we can always find the area under such well-behaved curves.

Theorem: Uniform Continuity of Continuous Functions on Compact Intervals

Continuous on Closed Interval implies “Extra Smooth” Continuity Before proving continuous functions are integrable, we need a powerful property: a function continuous on a closed and bounded interval (a “compact” set in $\mathbb{R}$) is not just continuous, but uniformly continuous. What’s uniform continuity? For regular continuity at a point $c$, for any $ϵ>0$, you need to find a $\delta >0$ such that if $x$ is $\delta$-close to $c$, then $f(x)$ is $ϵ$-close to $f(c)$. This $\delta$ can depend on $c$ and $ϵ$. For uniform continuity on an interval, for any $ϵ>0$, you can find a single $\delta >0$ that works for all pairs $x,y$ in the interval: if $|x-y|<\delta$, then $|f(x)-f(y)|<ϵ$. This $\delta$ depends only on $ϵ$, not on the specific location within the interval. It means the function’s “wiggliness” is controlled uniformly across the interval.

If $f:[a,b]\to \mathbb{R}$ is continuous on the closed and bounded interval $[a,b]$, then $f$ is uniformly continuous on $[a,b]$. This means:

For every $ϵ>0$, there exists a $\delta >0$ such that for all $x,y\in [a,b]$: If $|x-y|<\delta$, then $|f(x)-f(y)|<ϵ$.

(The crucial part is that this $\delta$ depends only on $ϵ$ and $f$, not on the specific points $x$ or $y$ in $[a,b]$).

Proof (by contradiction)

Assume $f$ is continuous on $[a,b]$ but not uniformly continuous on $[a,b]$. If $f$ is not uniformly continuous, then the definition of uniform continuity fails. This means there exists some ${ϵ}_0>0$ for which no $\delta >0$ works. That is, for every $\delta >0$, we can find a pair of points $x,y\in [a,b]$ such that $|x-y|<\delta$ but $|f(x)-f(y)|\ge {ϵ}_0$.

Let’s pick a sequence of ${\delta }_n=1/n$ (which goes to 0). For each $n\in {\mathbb{N}}^{\ast }$, we can find points $x_n,y_n\in [a,b]$ such that:

1. $|x_n-y_n|<1/n$
2. $|f(x_n)-f(y_n)|\ge {ϵ}_0$

Since $x_n\in [a,b]$ for all $n$, the sequence $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem, there exists a convergent subsequence $(x_{n_k})$ of $(x_n)$ that converges to some limit $x^{\ast }$. Since $[a,b]$ is a closed interval, $x^{\ast }$ must also be in $[a,b]$.

Now consider the corresponding subsequence $(y_{n_k})$. Since $|x_{n_k}-y_{n_k}|<1/n_k$ and $1/n_k\to 0$ as $k\to \infty$, it follows that $(y_{n_k})$ must converge to the same limit $x^{\ast }$. (Because $y_{n_k}=x_{n_k}+(y_{n_k}-x_{n_k})$, and $x_{n_k}\to x^{\ast }$ and $(y_{n_k}-x_{n_k})\to 0$).

Since $f$ is continuous on $[a,b]$, and $x^{\ast }\in [a,b]$, $f$ is continuous at $x^{\ast }$. Therefore: ${\lim }_{k\to \infty }f(x_{n_k})=f(x^{\ast })$ ${\lim }_{k\to \infty }f(y_{n_k})=f(x^{\ast })$

This implies that ${\lim }_{k\to \infty }(f(x_{n_k})-f(y_{n_k}))=f(x^{\ast })-f(x^{\ast })=0$.

However, this contradicts our construction where $|f(x_{n_k})-f(y_{n_k})|\ge {ϵ}_0>0$ for all $k$. This contradiction arises from the assumption that $f$ was not uniformly continuous. Therefore, the assumption must be false, and $f$ must be uniformly continuous on $[a,b]$.

Theorem: Continuous Functions are Riemann Integrable

The Big Payoff: Continuity Guarantees Integrability! This is a major result. If a function $f$ is continuous on a closed, bounded interval $[a,b]$, then it is always Riemann integrable on $[a,b]$. This is why we can confidently talk about “the area under the curve” for most functions we encounter in calculus.

If $f:[a,b]\to \mathbb{R}$ is continuous, then $f$ is Riemann integrable on $[a,b]$.

Proof: We will use Riemann’s Criterion (Theorem 5.1.8): show that for any $ϵ>0$, there exists $\delta >0$ such that for all partitions $P$ with mesh $\delta (P)<\delta$, we have $S(f,P)-s(f,P)<ϵ$.

1. Boundedness: Since $f$ is continuous on the closed, bounded interval $[a,b]$, by the Extreme Value Theorem (Min-Max Theorem), $f$ is bounded on $[a,b]$. This is a prerequisite for Riemann integrability.

2. Uniform Continuity: Since $f$ is continuous on $[a,b]$, by the theorem just proven, $f$ is uniformly continuous on $[a,b]$. Let $ϵ>0$ be given. Because of uniform continuity, for the value ${ϵ}^{\prime }=\frac{ϵ}{b-a}$ (assuming $b-a>0$; if $b=a$, the interval has zero length and the integral is 0, so $f$ is trivially integrable), there exists a $\delta >0$ such that for all $x,y\in [a,b]$: If $|x-y|<\delta$, then $|f(x)-f(y)|<{ϵ}^{\prime }=\frac{ϵ}{b-a}$.

   Now, choose any partition $P=\{a=x_0,x_1,\dots ,x_n=b\}$ of $[a,b]$ such that its mesh $\delta (P)<\delta$. This means that for every subinterval $I_i=[x_{i-1},x_i]$, its length $x_i-x_{i-1}<\delta$. On each such closed subinterval $I_i$, $f$ is continuous. Thus, by the Extreme Value Theorem, $f$ attains its minimum value $m_i$ and its maximum value $M_i$ within $I_i$. Let $y_i^{\ast },x_i^{\ast }\in I_i$ be points such that $f(y_i^{\ast })=m_i$ and $f(x_i^{\ast })=M_i$. Since $x_i^{\ast }$ and $y_i^{\ast }$ are both in $I_i$, the distance between them $|x_i^{\ast }-y_i^{\ast }|\le (x_i-x_{i-1})<\delta$. Because $|x_i^{\ast }-y_i^{\ast }|<\delta$, by uniform continuity, we have: $M_i-m_i=f(x_i^{\ast })-f(y_i^{\ast })=|f(x_i^{\ast })-f(y_i^{\ast })|<{ϵ}^{\prime }=\frac{ϵ}{b-a}$. (Note: $M_i\ge m_i$, so $M_i-m_i\ge 0$).

   Now consider the difference between the upper and lower Darboux sums for this partition $P$: $S(f,P)-s(f,P)={\sum }_{i=1}^n{M}_i(x_i-x_{i-1})-{\sum }_{i=1}^n{m}_i(x_i-x_{i-1})$ $={\sum }_{i=1}^n(M_i-m_i)(x_i-x_{i-1})$ Since $M_i-m_i<\frac{ϵ}{b-a}$ for all $i$: $S(f,P)-s(f,P)<{\sum }_{i=1}^n\left(\frac{ϵ}{b-a}\right)(x_i-x_{i-1})$ $=\frac{ϵ}{b-a}{\sum }_{i=1}^n(x_i-x_{i-1})$ The sum ${\sum }_{i=1}^n(x_i-x_{i-1})$ is the sum of the lengths of all subintervals, which is $(x_1-x_0)+(x_2-x_1)+\cdots +(x_n-x_{n-1})=x_n-x_0=b-a$. $S(f,P)-s(f,P)<\frac{ϵ}{b-a}(b-a)=ϵ$ So, for every $ϵ>0$, we have found a $\delta >0$ (from uniform continuity) such that for any partition $P$ with mesh $\delta (P)<\delta$, we have $S(f,P)-s(f,P)<ϵ$. By Theorem 5.1.8 (Riemann’s Criterion for Integrability), $f$ is Riemann integrable on $[a,b]$. $\square$