## Lebesgue on Riemann Integration

When is a bounded, real-valued function $f:[a,b]\rightarrow\mathbb{R}$ Riemann integrable? I ask this question for a few reasons. The first is because I assigned a couple exercises on the my students’ take-home final exam introducing ‘measure zero’ and the types of discontinuities of functions. The second is that I was reading the statement of Weyl equidistribution theorem, where the condition is Riemann integrable, as opposed to Lebesgue integrable. In fact, the theorem is false if you use the latter. The third and last is that I have known the answer for some time, but I have never worked through the proof in full detail. Well, I plan to change that with today’s post.

Theorem 1. (Lebesgue) A bounded function $f: [a,b]\rightarrow\mathbb{R}$ is Riemann integrable if and only if the set of discontinuities of $f$ is of measure zero.

It turns out that a bounded function is Riemann integrable if and only if it is continuous and its discontinuities form a set of measure zero. The French mathematician Henri Lebesgue proved this result over a century ago. It is important to not confuse this ‘Lebesgue’s theorem’ with other results about the Lebesgue integral, a more powerful integral based on measure theory. However, the limitations of the Riemann integral shown by Lebesgue motivate the study of his integral.

Indeed, we would think that a function zero almost everywhere (a.e.) would be Riemann integrable with integral zero. This is not the case. Consider the Dirichlet function

$\displaystyle f(x):=\begin{cases}1&{x\in\mathbb{Q}}\\ 0&{x\in\mathbb{R}\setminus\mathbb{Q}}\end{cases}$

which is zero a.e. However, $f$ is not Riemann integrable on any compact interval $[a,b]$. Given any partition $P=\left\{a=x_{0}, the upper Riemann sum

$\displaystyle\overline{S}(f;P):=\sum_{j=1}^{n}\underbrace{\sup_{t\in[x_{j-1},x_{j}]}f(t)}_{M_{j}}\underbrace{(x_{j}-x_{j-1})}_{\Delta x_{j}}$

is $1$ while the lower Riemann sum

$\displaystyle\underline{S}(f;P):=\sum_{j=1}^{n}\underbrace{\inf_{t\in[x_{j-1},x_{j}]}f(t)}_{m_{j}}\underbrace{(x_{j}-x_{j-1})}_{\Delta x_{j}}$

is $0$ by the density of the rationals in the reals.

Enough with the motivation, let’s get down to the details. To prove Lebesgue’s theorem, we will need to define the oscillation of a bounded function $f$ at a point $x$, which quantifies how much $f$ varies in an arbitrarily small interval about $x$. Precisely, the oscillation of $f:[a,b]\rightarrow\mathbb{C}$ at the point $x\in[a,b]$ is

$\displaystyle\text{osc}(f;x):=\lim_{h\rightarrow0^{+}}\sup_{x',x''\in(x-h,x+h)}\left|f(x')-f(x'')\right|$

Note that the supremum is well-defined since $f$ is bounded. The oscillation of $f$ at a point $x$ is a useful notion, since it compactly encodes the continuity of a function.

Lemma 2. $f:[a,b]\rightarrow\mathbb{C}$ is continuous at a point $x\in[a,b]$ if and only if $\text{osc}(f;x)=0$.

$\displaystyle\forall x'\in(x-\delta,x+\delta) \ \left|f(x)-f(x')\right|<\dfrac{\varepsilon}{2}$,

which implies by the triangle inequality that

$\displaystyle\forall x',x''\in(x-\delta,x+\delta) \ \left|f(x')-f(x'')\right|<\varepsilon\Longrightarrow\sup_{x',x''\in(x-\delta,x+\delta)}\left|f(x')-f(x'')\right|\leq\varepsilon$

Letting $\delta\rightarrow0$, we see that $\text{osc}(f;p)\leq\varepsilon$. Since $\varepsilon$ was arbitrary, we conclude that $\text{osc}(f;p)=0$.

The other direction of the proof essentially follows from retracing our steps above, and I leave it as an exercise to the reader. $\Box$

Lemma 3. Let $f: [a,b]\rightarrow\mathbb{C}$ be a bounded function. Then for any $s>0$,

$\displaystyle A_{s}:=\left\{x\in[a,b]:\text{osc}(f;x)\geq s\right\}$

is compact.

Proof. Since $[a,b]$ is compact, it suffices to show that $A_{s}$ is a closed subset of $[a,b]$, or $A_{s}^{c}$ is open in $[a,b]$. Let $x_{0}\in A_{s}^{c}$, so that $\text{osc}(f;x_{0}) =: t . Set $\varepsilon:=\frac{s-t}{2}$. By definition of oscillation, there exists a constant $\delta=\delta(\epsilon)>0$ such that $0 implies that

$\displaystyle\left|t-\sup_{x',x''\in (x_{0}-h,x_{0}+h)}\left|f(x')-f(x'')\right|\right|<\varepsilon$

By the triangle inequality,

$\displaystyle\sup_{x',x''\in (x_{0}-\delta,x_{0}+\delta)}\left|f(x')-f(x'')\right|+t<\frac{s-t}{2}+t=\frac{s+t}{2},

So, for any $y_{0}\in (x_{0}-\delta,x_{0}+\delta)$, we can take $h>0$ sufficiently small so that $(y_{0}-h,y_{0}+h)\subset(x_{0}-\delta,x_{0}+\delta)$ and therefore

$\displaystyle\text{osc}(f;y_{0})=\lim_{h\rightarrow0^{+}}\sup_{x',x''\in (y_{0}-h,y_{0}+h)}\left|f(x')-f(x'')\right|\leq\frac{s+t}{2}

Equivalently, $(x_{0}-\delta,x_{0}+\delta)\subset A_{s}^{c}$. $\Box$

Let $M$ and $m$ be the supremum and infimum of $f$, respectively, on $[a,b]$. Denote the set of discontinuities of $f$ by $A$, and note that

$\displaystyle A:=\left\{x\in[a,b]:\text{osc}(f;x)>0\right\}$,

which can be represented as the countable union

$\displaystyle A=\bigcup_{k=1}^{\infty}A_{k},\indent A_{k}:=\left\{x\in[a,b]:\text{osc}(f;x)\geq\frac{1}{k}\right\}$

First, suppose that $f$ is Riemann integrable. Then, for any $\varepsilon>0$ given, there exists a partition

$\displaystyle P:=\left\{a=x_{0}

such that

$\displaystyle\overline{S}(f;{P})-\underline{S}(f;{P})=\sum_{j=1}^{n}\left[M_{j}-m_{j}\right]\underbrace{(x_{j-1}-x_{j})}_{\Delta x_{j}}<\dfrac{\varepsilon}{k}$

If $x\in(x_{j-1},x_{j})\cap A_{k}$, for some $j\in\left\{1,\cdots,n\right\}$ and $k\in\mathbb{N}$, then

$\displaystyle(M_{j}-m_{j})\Delta x_{j}\geq\dfrac{1}{k}\Delta x_{j}$

Hence, if $I=\left\{1\leq j\leq n: A_{k}\cap(x_{j-1},x_{j})\neq\emptyset\right\}$, then

$\displaystyle\dfrac{\varepsilon}{k}>\sum_{j=1}^{n}[M_{j}-m_{j}]\Delta x_{j}\geq\sum_{j\in I}[M_{j}-m_{j}]\Delta x_{j}\geq\dfrac{1}{k}\sum_{j\in I}\Delta x_{j}$

Multiplying both sides of the inequality by $k$, we see that $\sum_{j\in I}\Delta x_{j}<\varepsilon$. The intervals $(x_{j-1},x_{j})$, for $j\in I$, form an open cover of $A_{k}\setminus\left\{x_{0},\cdots,x_{n}\right\}$, which is equal in measure to $A_{k}$. Hence,

$\displaystyle\left|A_{k}\right|\leq\sum_{j\in I}\Delta x_{j}<\varepsilon,$

which implies that $\left|A_{k}\right|=0$. Since the countable union of null sets is again a null set, we conclude that $\left|A\right|=0$.

Now suppose that $\left|A\right|=0$. Since $A_{s}\subset A$, we see that $\left|A_{s}\right|=0$, for all $s>0$. Hence, for $\varepsilon>0$ be given, there exists a countable collection of open intervals $\left\{I_{k}\right\}_{k=1}^{\infty}$ such that

$\displaystyle A_{\frac{\varepsilon}{2(b-a)}}\subset\bigcup_{k=1}^{\infty}I_{k},\indent\sum_{k=1}^{\infty}\left|I_{k}\right|<\dfrac{\varepsilon}{2(M-m)}$

The preceding lemma implies that $A_{\varepsilon}$ is compact, hence there exists a finite subcover $\left\{I_{1},\cdots,I_{N}\right\}$ of $A_{\varepsilon}$. If $x\in[a,b]\setminus\bigcup_{j=1}^{N}I_{j}$, then $\text{osc}(f;x)<\varepsilon$. So for such an $x$, there exists $\delta_{x}>0$ such that for all $x',x''\in (x-\delta_{x},x+\delta_{x})$, $\left|f(x')-f(x'')\right|<\frac{\varepsilon}{2(b-a)}$. The collection of intervals

$\displaystyle\left\{(x-\delta_{x},x+\delta_{x}): x\in [a,b]\setminus\bigcup_{j=1}^{N}I_{j}\right\}$

forms an open cover of $[a,b]\setminus\bigcup_{j=1}^{N}I_{j}$, which is compact. Hence, there exists a finite subcover

$\displaystyle\left\{(\tilde{x}_{1}-\delta_{1},\tilde{x}_{1}+\delta_{1}),\cdots,(\tilde{x}_{k}-\delta_{k},\tilde{x}_{k}+\delta_{k})\right\}$

This collection of intervals together with the intervals $I_{1},\cdots,I_{N}$ form an open cover of $[a,b]$. We want to use this open cover to obtain a partition $P$ of $[a,b]$ such that $\overline{S}(f;P)-\underline{S}(f;P)<\varepsilon$. In particular, we want the partition $a=x_{0} to have the property that any interval $[x_{j-1},x_{j}]$ is contained entirely in an element of the open cover, for which we will need Lebesgue’s number lemma. We postpone the proof of the lemma until the end in favor of completing the proof of Theorem 1.

Lemma 4. (Lebesgue) If $(X,d)$ is a compact metric space and $\mathcal{A}$ is an open cover of $X$, then there exists a $\delta>0$ such that, for every subset $C\subset X$ with $\text{diam } C<\delta$, there exists some $A\in\mathcal{A}$ such that $C\subset A$.

Divide $P$ into two sets $P_{1}$ and $P_{2}$, where

$\displaystyle P_{1}:=\left\{x_{j}\in P:[x_{j-1},x_{j}]\subset I_{k}\text{ for some }k_{j}\right\},\indent P_{2}:=P\setminus C_{1}$

For points in $P_{1}$, the oscillation can be ‘big’, but the length of the interval $[x_{j-1}-x_{j}]$ is very small. For points in $P_{2}$, the oscillation is sufficiently ‘small’, and the total length of the intervals is at most $b-a$.

$\begin{array}{lcl}\displaystyle\overline{S}(f;P)-\underline{S}(f;P)&=&\displaystyle\sum_{x_{j}\in C_{1}}[M_{j}-m_{j}]\Delta x_{j}+\sum_{x_{j}\in C_{2}}[M_{j}-m_{j}]\Delta x_{j}\\[.9 em]&\leq&\displaystyle(M-m)\sum_{x_{j}\in C_{1}}\left|I_{k_{j}}\right|+\dfrac{\varepsilon}{2(b-a)}\sum_{x_{j}\in C_{2}}\Delta x_{j}\\[.9 em]&\leq&\displaystyle(M-m)\cdot\dfrac{\varepsilon}{2(M-m)}+\dfrac{\varepsilon}{2(b-a)}\cdot(b-a)\\[.9 em]&=&\displaystyle\varepsilon\end{array}$

Proof of Lebesgue’s Number Lemma. We may assume that $X\notin\mathcal{A}$, otherwise any $\delta>0$ works. Since $X$ is compact, there exist finitely many elements $A_{1},\cdots,A_{n}$ in $\mathcal{A}$ such that $X\subset\bigcup_{j=1}^{n}A_{j}$. For each $j=1,\cdots,n$, define $B_{j}:=X\setminus A_{j}$. Define a real-valued function $f$ on $X$ by

$\displaystyle f(x):=\dfrac{1}{n}\sum_{j=1}^{n}d(x,B_{j}),\indent\forall x\in X$

I claim that $f(x)>0$ for all $x\in X$. Indeed, if $x\in A_{j}$, then we can take $\varepsilon>0$ sufficiently small so that the open ball $B(x;\varepsilon)\subset A_{j}$. Hence, $d(x,B_{j})\geq\varepsilon$ and $f(x)\geq\frac{\varepsilon}{n}$.

$f$ is continuous and $X$ is compact, so by the extreme value theorem, $f$ attains its minimum $\delta>0$ on $X$. Suppose we are given a set $C\subset X$ such that $\text{diam }C<\delta$. For any point $x_{0}\in C$, the ball $B(x_{0};\delta)$ contains the set $C$, by definition of diameter. If $m$ denotes the index of a set satisfying $d(x_{0},B_{m})=\max\left\{d(x_{0},B_{j}):1\leq j\leq n\right\}$, then

$\displaystyle\delta\leq f(x_{0})=\dfrac{1}{n}\sum_{j=1}^{n}d(x_{0},B_{j})\leq d(x_{0},B_{m})$,

which implies that $C\subset B(x_{0};\delta)\subset X\setminus B_{m}=A_{m}$. $\Box$