This blog was originally written in Chinese and translated by Deepseek-v4-flash. Though I've reviewed and polished the translation, feel free to contact me if any phrasing seems unnatural.

Abstract: Part 1 lists three necessary and sufficient criteria for Riemann integrability that are covered in most textbooks; Part 2 covers the preparatory knowledge required before the proof; Part 3 completes the proof.

In standard analysis textbooks, the Lebesgue theorem is only stated as a necessary and sufficient condition for Riemann integrability without providing its proof. This article proves the Lebesgue theorem without involving higher-level analysis (such as real variable functions).

Note: The following procedure does not consider the case of endpoints (for example, the oscillation function at endpoints is defined by one-sided intervals); in fact, the proof method is entirely similar, only requiring a change of notation.

Part 1

First, we list three necessary and sufficient criteria for Riemann integrability:

(The following defaults: $\Delta$ is a partition of $[a,b]$ with $x_0=a<\dots<b=x_n$, $\Delta x_i=x_i-x_{i-1}$, $\lambda(\Delta)=\max\{\Delta x_i\}$, and $w_i$ is the oscillation on $[x_{i-1},x_i]$.)

Theorem 1: The necessary and sufficient condition for $f(x)\in R[a,b]$ is:

$\forall\varepsilon>0,\exists\delta,\forall\Delta$, as long as $\lambda(\Delta)<\delta$, we have $\sum_{i=1}^n w_i\Delta x_i<\varepsilon$.
$\forall\varepsilon>0,\exists\Delta$, satisfying $\sum_{i=1}^n w_i\Delta x_i<\varepsilon$.
$\forall\varepsilon>0,\forall\sigma>0,\exists\Delta$, such that the sum of the lengths of subintervals whose oscillation exceeds $\varepsilon$ is less than $\sigma$.

These three theorems are covered and proved in standard textbooks, so we will not repeat their proofs here.

Part 2

Before formally beginning the proof, we first introduce the concept of the oscillation function of a function on the real numbers:

Definition: Let $w_\delta$ be the oscillation of $f(x)$ on $[x-\delta,x+\delta]$. Define $w(x)=\lim\limits_{\delta\to0^+}w_\delta$.

Here, $w_\delta$ can also be written as $\sup\limits_{x_1,x_2\in[x-\delta,x+\delta]}\{f(x_1)-f(x_2)\}=\sup\limits_{x\in[x-\delta,x+\delta]}f(x)-\inf\limits_{x\in[x-\delta,x+\delta]}f(x)$.

Theorem 2: $f$ is continuous at $x_0$ if and only if $w(x_0)=0$.

This is very easy to prove.

Not every function has an oscillation function defined everywhere. For example, take $f(x)=\frac{1}{x}$ and define $f(0)=0$; then the oscillation function is undefined at 0. However, integrable functions are all bounded, which leads us to consider whether the oscillation function of a bounded function is defined everywhere.

Proposition: The oscillation function of a bounded function is defined everywhere.

Proof:

Observe that $w_\delta=\sup\limits_{x\in[x-\delta,x+\delta]}f(x)-\inf\limits_{x\in[x-\delta,x+\delta]}f(x)$. The term $\sup\limits_{x\in[x-\delta,x+\delta]}f(x)$ decreases as $\delta$ decreases (i.e., it is monotonically increasing in $\delta$), and $\inf\limits_{x\in[x-\delta,x+\delta]}f(x)$ increases as $\delta$ decreases (i.e., it is monotonically decreasing in $\delta$). Consequently, $w_\delta$ is a monotonically increasing function of $\delta$. Since the function is bounded, the oscillation is also bounded, and by the monotone bounded convergence principle, $w(x)$ always exists.

With the introduction of the oscillation function, a small part of our work is done. Next, we present an intermediate step for proving the Lebesgue criterion:

Theorem 3 (du Bois-Reymond criterion): $f\in R[a,b]$ if and only if $f(x)$ is bounded on $[a,b]$ and for every $\varepsilon>0,\sigma>0$, the set $\{x\in[a,b]:w(x)\geq\varepsilon\}$ can be covered by finitely many open intervals whose total length is $\leq\sigma$.

Proof:

We consider proving the equivalence of this theorem with the third condition of Theorem 1, since their forms are quite similar.

”$\Leftarrow$”: Given that $f(x)$ is bounded on $[a,b]$, and for every $\varepsilon>0,\sigma>0$, the set $\{x\in[a,b]:w(x)\geq\varepsilon\}$ can be covered by finitely many open intervals whose total length is $\leq\sigma$.

Consider the definition of the oscillation function: $w_\delta$ decreases as $\delta$ decreases, so for any $\delta_0>0$, the oscillation of $f(x)$ on $[x_0-\delta_0,x_0+\delta_0]$ satisfies $w\geq\lim\limits_{\delta\to0}w_\delta=w(x_0)$. By the condition of the theorem, we thus have $w\geq\varepsilon$. This means that any interval containing a point where the original oscillation exceeds $\varepsilon$ also has oscillation greater than $\varepsilon$.

For any such point $x_0$, suppose it is covered by an open interval $(c,d)$. Then we can take $\delta_0=\frac{1}{2}\min\{x_0-c,d-x_0\}$ so that $[x_0-\delta_0,x_0+\delta_0]$ is also covered by $(c,d)$. Consequently, the union of all intervals containing these points can also be covered by finitely many open intervals whose total length is $\leq\sigma$.

For points where the oscillation is less than $\varepsilon$, by the order-preserving property, there exists $\delta_1>0$ such that the oscillation of $f(x)$ on $[x_0-\delta_1,x_0+\delta_1]$ is less than $\varepsilon$.

Since there are only finitely many open intervals, we can partition the remaining part of $[a,b]$ into a union of closed intervals. Within each closed interval, for every point, take the aforementioned neighborhood (excluding endpoints) $U(x_0,\delta_1)$, thereby covering this closed interval. By the Heine-Borel theorem, there exists $m>0$ such that these $m$ open intervals cover this closed interval; supplementing the endpoints, these $m$ closed intervals also cover this closed interval. Thus, we obtain at most $2m$ partition points of closed intervals, where any closed interval formed by two adjacent points is covered by at least one of the original closed intervals, and consequently its oscillation is also less than $\varepsilon$.

Therefore, suppose there are $K$ open intervals, which then form at most $K+1$ closed intervals. Each of these can be covered by $m_1,\dots,m_K$ closed intervals respectively, and by these partition points (at most $2(m_1+\cdots+m_K)$ in total), we obtain a partition that satisfies the third condition of Theorem 1.
”$\Rightarrow$”: It is obvious that $f$ is bounded on $[a,b]$.

Given that for every $\varepsilon>0,\sigma>0$, there exists a partition $\Delta$ such that the sum of lengths of subintervals whose oscillation exceeds $\varepsilon$ is $\sum_{i=1}^k\Delta x_i<\frac{\sigma}{2}$.

Apart from the partition points themselves (since any neighborhood of a partition point spans two subintervals and cannot be covered by a single small subinterval), by the above discussion, the oscillation at a point cannot exceed the oscillation of the subinterval containing it. Thus, within subintervals where the oscillation is less than $\varepsilon$, there are at most finitely many points (say $N$ points) whose oscillation exceeds $\varepsilon$.

Suppose there are $K$ subintervals with oscillation $\geq\varepsilon$. Let any such subinterval be $[x_{i-1},x_i]$, and take $(x_{i-1}-\frac{\sigma}{2^{1+i}},x_i+\frac{\sigma}{2^{1+i}})$. For the remaining $N$ points, take their neighborhoods with radii successively $\frac{\sigma}{2^{N+2}},\dots,\frac{\sigma}{2^{N+K+2}}$. Then the total length of these $N+K$ open intervals is $\leq \frac{\sigma}{2}+\left(\frac{\sigma}{4}+\cdots+\frac{\sigma}{2^{N+K+1}}\right)<\sigma$.

This completes the proof.

Part 3

Now we arrive at the final proof:

Theorem: $f\in R[a,b]$ if and only if $f$ is bounded and its set of discontinuity points is a null set (i.e., $f$ is continuous almost everywhere on $[a,b]$).

Here, the concept of a null set in $\mathbb{R}$ is: Let $S$ be a subset of the real numbers. If for every $\varepsilon>0$, there exists a family of at most countably many open intervals $I_j=(a_j,b_j)\;(j=1,2,\dots)$ such that the sum of their lengths $\sum_{j=1}^n(b_j-a_j)$ (where $n$ can be infinite) is less than $\varepsilon$, then $S$ is called a null set.

Proof:

In fact, let us compare Theorem 3 (du Bois-Reymond criterion) with the Lebesgue criterion:

Theorem 3 (du Bois-Reymond criterion): $f(x)$ is bounded on $[a,b]$, and for every $\eta>0,\sigma>0$, the set $\{x\in[a,b]:w(x)\geq\eta\}$ can be covered by finitely many open intervals whose total length is $\leq\sigma$.

Lebesgue criterion: $f(x)$ is bounded on $[a,b]$, and for every $\varepsilon>0$, the set $\{x\in[a,b]:w(x)\neq0\}$ can be covered by at most countably many open intervals whose total length is less than $\varepsilon$.

As can be seen, the two are already very close in form, so we consider proving their equivalence. The difficulty of the proof lies in the transition between “finite” and “infinite”: one comes from the transition between $\forall\eta>0$ and $0$, and the other from the transition between finite and at most countable.

”$\Rightarrow$”: Given that the function is integrable, i.e., we prove the Lebesgue criterion from Theorem 3 (du Bois-Reymond criterion).

Observe that Theorem 3 states: for every $\eta>0$, an integrable function satisfies that $\{x\in[a,b]:w(x)\geq\eta\}$ is always a null set.

By Theorem 2, points where the oscillation function is non-zero are necessarily discontinuity points. Consider taking $\eta=\frac{1}{n}$. Then for any discontinuity point $x_0$ with $w(x_0)\neq0$, there exists $N$ such that $w(x_0)\in\left[\frac{1}{N+1},\frac{1}{N}\right]$ (in addition, there may be a set of points where the oscillation exceeds $1$; by the theorem this is also a null set). By the above assertion, it follows that $\left\{x\in[a,b]:w(x)\in\left[\frac{1}{N+1},\frac{1}{N}\right]\right\}$ is a null set. Consequently, the set of discontinuity points is a union of countably many null sets; the union of countably many null sets is still a null set, so the set of discontinuity points is also a null set.

”$\Leftarrow$”: Inferring integrability from the Lebesgue criterion:

For a continuity point of $f(x)$, by the definition of continuity (to avoid confusion, we use $\xi$ in place of $\varepsilon$ here): $\forall\xi>0,\exists\delta>0$ such that $\forall x\in U(x_0,\delta)$, $$

f(x)-f(x_0)

<\xi$. Take$\xi=\frac{\eta}{4}$; then within$U(x_0,\delta)$, the oscillation of$f(x)$is no more than$2\xi=\frac{\eta}{2}$. For all continuity points, take such open intervals and denote them as class$(I)$intervals. For discontinuity points, by the null set property, take open intervals that cover them and denote them as class$(II)$intervals. Clearly, every point either belongs to a class$(I)$interval, or belongs to a class$(II)$interval, or is covered by both types of intervals. Thus, all points together form a family of open intervals covering the entire closed interval. By the Heine-Borel theorem, there exists$p>0$such that the closed interval is covered by these$p$$ open intervals.

Using the partition points of these $p$ open intervals, take a partition $\Delta:a=x_0<\cdots<x_n=b$ of the closed interval. For any open interval between two adjacent partition points, it is either contained in a class $(I)$ interval, meaning its oscillation does not exceed $\frac{\eta}{2}$, hence less than $\eta$; or it is not contained in any class $(I)$ interval, in which case it must belong to a class $(II)$ interval, and for every $\varepsilon>0$, the class $(II)$ intervals can be covered by open intervals whose total length is less than $\varepsilon$. As a subset, the total length of the open intervals here is also less than $\varepsilon$. Take $\varepsilon=\frac{\sigma}{2}$. Then among the open intervals formed by adjacent partition points, those points whose oscillation exceeds $\eta$ can be covered by finitely many open intervals (say $k$ of them), with total length less than $\frac{\sigma}{2}$.

Now only the partition points themselves, i.e., the endpoints of the closed subintervals, remain to be discussed. Since the partition points are finite, there obviously exist at most $n+1$ open intervals covering these partition points, with total length less than $\frac{\sigma}{2}$. Consequently, the set of points where the oscillation exceeds $\eta$ can be covered by finitely many open intervals (at most $n+k+1$). Thus, we have proved that for every $\eta>0,\sigma>0$, the set $\{x\in[a,b]:w(x)\geq\eta\}$ can be covered by finitely many open intervals whose total length is $\leq\sigma$.

Q.E.D.