Continuous Functions Have Lebesgue Points

Lebesgue Points and Local Oscillation

At a point where a function is continuous, its values throughout a sufficiently small ball are close to its value at the center. Averaging that pointwise difference over the ball preserves the same bound. This simple observation gives a precise form of the Lebesgue-point property, with no exceptional points: continuity makes every point a Lebesgue point.

Throughout, $m$ denotes Lebesgue measure on $\mathbb{R}^n$, and $f:\mathbb{R}^n\to\mathbb{R}$ is continuous. For $x\in\mathbb{R}^n$ and $r>0$, let $B(x,r)=\{y\in\mathbb{R}^n:|y-x|<r\}$. Write $v_n=m(B(0,1))$, so $0<m(B(x,r))=v_n r^n<\infty$. Continuity implies that $f$ is bounded on every compact set; hence $f$ is locally integrable, meaning that its absolute value has finite integral on every compact set.

Definition: A point $x\in\mathbb{R}^n$ is a Lebesgue point of a locally integrable, real-valued function $f$, with the chosen value $f(x)$, if $$ \lim_{r\downarrow0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y)=0. $$ This is a limit of averages of absolute differences, not signed differences. For continuous $f$, the definition applies at every $x$, using its actual pointwise value.

The key quantitative object is a modulus of continuity restricted to a compact set. It measures the largest change in $f$ between two points of that set whose distance is at most a prescribed amount.

Definition: If $K\subset\mathbb{R}^n$ is a nonempty compact set and $\delta\ge0$, define $$ \omega_{f,K}(\delta)=\sup\{|f(u)-f(v)|:u,v\in K,\ |u-v|\le\delta\}. $$ The supremum is finite because $f$ is bounded on $K$. In particular, $\omega_{f,K}(0)=0$.

Compact Neighborhoods Give an Explicit Bound

Lemma: If $f:\mathbb{R}^n\to\mathbb{R}$ is continuous and $K\subset\mathbb{R}^n$ is nonempty and compact, then $\omega_{f,K}(\delta)\to0$ as $\delta\downarrow0$.

Proof. Suppose instead that the limit is not zero. Since $\omega_{f,K}(\delta)$ is nondecreasing in $\delta$, there would be an $\varepsilon_0>0$ and a sequence $\delta_j\downarrow0$ such that $\omega_{f,K}(\delta_j)\ge\varepsilon_0$ for every $j$. By the definition of supremum, choose $u_j,v_j\in K$ with $|u_j-v_j|\le\delta_j$ and $|f(u_j)-f(v_j)|>\varepsilon_0/2$. Compactness gives a subsequence of $(u_j)$ converging to some $u\in K$. Along that subsequence, $|v_j-u|\le|v_j-u_j|+|u_j-u|\to0$, so $v_j\to u$ as well. Continuity at $u$ then gives $f(u_j)\to f(u)$ and $f(v_j)\to f(u)$, contradicting $|f(u_j)-f(v_j)|>\varepsilon_0/2$. Therefore $\omega_{f,K}(\delta)\to0$. $\square$

Theorem: Let $f:\mathbb{R}^n\to\mathbb{R}$ be continuous. For every $x\in\mathbb{R}^n$, there is a compact neighborhood $K$ of $x$ such that for every $r>0$ with $B(x,r)\subset K$, $$ \frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y) \le \omega_{f,K}(r). $$ Consequently, every $x\in\mathbb{R}^n$ is a Lebesgue point of $f$.

Proof. Fix $x\in\mathbb{R}^n$ and take $K=\overline{B(x,1)}$, which is compact and contains a neighborhood of $x$. If $B(x,r)\subset K$, then for every $y\in B(x,r)$ both $x,y\in K$ and $|y-x|<r$. The definition of the modulus therefore gives $|f(y)-f(x)|\le\omega_{f,K}(r)$. Integrating this pointwise inequality over $B(x,r)$, then dividing by its positive finite measure, proves the displayed bound. By the lemma, $\omega_{f,K}(r)\to0$ as $r\downarrow0$; the average is nonnegative and bounded above by this quantity for all sufficiently small $r$, so it tends to zero. This is exactly the Lebesgue-point condition. $\square$

The bound is local: it does not require $f$ to be uniformly continuous on all of $\mathbb{R}^n$. The compact set can be chosen around the point under consideration, and the modulus on that neighborhood controls every sufficiently small ball there.

Worked Example: The Absolute-Value Function on the Line

Let $f(y)=|y|$ on $\mathbb{R}$, with Lebesgue measure, and find the average oscillation at $x=0$. For $r>0$, $m((-r,r))=2r$, and $f(0)=0$. Thus

\frac{1}{m((-r,r))}\int_{-r}^{r}|f(y)-f(0)|\,dy =\frac{1}{2r}\int_{-r}^{r}|y|\,dy =\frac{1}{2r}\left(2\int_0^r y\,dy\right) =\frac{r}{2}.

This tends to zero, so $0$ is a Lebesgue point. In fact, on any compact interval $K$, $||u|-|v||\le|u-v|$; hence $\omega_{f,K}(r)\le r$, and the exact average $r/2$ is consistent with the general bound. Check: substituting $r=2$ into the integral gives $\int_{-2}^{2}|y|\,dy=2\int_0^2y\,dy=4$; dividing by $m((-2,2))=4$ gives $1=2/2$.

Worked Example: A Quadratic Function in Euclidean Space

For $f(y)=|y|^2$ on $\mathbb{R}^n$, compute the average oscillation at the origin. Since $f(0)=0$, the integrand is $|y|^2$. Let $v_n=m(B(0,1))$. Using polar coordinates, for $r>0$,

\int_{B(0,r)}|y|^2\,dm(y) =n v_n\int_0^r \rho^2\rho^{n-1}\,d\rho =n v_n\frac{r^{n+2}}{n+2}.

Here the radial factor $n v_n\rho^{n-1}$ is the surface-area factor corresponding to the volume $v_n\rho^n$ of a radius-$\rho$ ball. Dividing by $m(B(0,r))=v_n r^n$ gives

\frac{1}{m(B(0,r))}\int_{B(0,r)}|f(y)-f(0)|\,dm(y) =\frac{n}{n+2}r^2\longrightarrow0.

Thus the origin is a Lebesgue point. Check: when $n=1$, the formula gives $r^2/3$. Directly, $\frac{1}{2r}\int_{-r}^r y^2\,dy=\frac{1}{2r}\cdot\frac{2r^3}{3}=r^2/3$, as required.

Worked Example: The Sine Function at the Origin

Let $f(y)=\sin y$ on $\mathbb{R}$, with $x=0$. For $0<r\le\pi$, $|\sin y|=\sin |y|$ on $[-r,r]$, and $\sin y\ge0$ on $[0,r]$. Therefore

\frac{1}{2r}\int_{-r}^{r}|\sin y-\sin 0|\,dy =\frac{1}{r}\int_0^r\sin y\,dy =\frac{1-\cos r}{r}.

The elementary inequality $1-\cos r\le r^2/2$ gives an upper bound $r/2$, which tends to zero. Hence $0$ is a Lebesgue point. Check: at $r=\pi/2$, the exact average is $(1-\cos(\pi/2))/(\pi/2)=2/\pi$, and the bound is $(\pi/2)/2=\pi/4$; indeed $2/\pi\le\pi/4$, equivalently $8\le\pi^2$.

Uniform Control over Compact Sets

The same reasoning gives a bound that works simultaneously for all centers in a fixed compact set. To ensure the averaging balls remain in a compact region, for a nonempty compact $K\subset\mathbb{R}^n$ and $R>0$ define $K_R=\{y:\operatorname{dist}(y,K)\le R\}$, where $\operatorname{dist}(y,K)=\inf_{z\in K}|y-z|$. The set $K_R$ is closed and bounded, hence compact.

Corollary: If $f:\mathbb{R}^n\to\mathbb{R}$ is continuous, $K\subset\mathbb{R}^n$ is nonempty and compact, and $R>0$, then for every $x\in K$ and $0<r\le R$, $$ \frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y) \le\omega_{f,K_R}(r). $$ In particular, the supremum of these average oscillations over all $x\in K$ is bounded by the same quantity, which tends to zero as $r\downarrow0$.

Proof. Fix $x\in K$ and $0<r\le R$. The center $x$ belongs to $K_R$. For $y\in B(x,r)$, $\operatorname{dist}(y,K)\le|y-x|<r\le R$, so $y\in K_R$ as well. Also $|y-x|<r$. Thus $|f(y)-f(x)|\le\omega_{f,K_R}(r)$ throughout the ball. Integrating and dividing by its positive finite measure proves the inequality. Its right-hand side is independent of $x$, so it bounds the supremum over $K$. The compact-set lemma gives $\omega_{f,K_R}(r)\to0$. $\square$

This uniform version is useful when the center varies over a compact region: continuity supplies one common bound rather than a separate continuity radius for every point. It does not say that a continuous function on all of $\mathbb{R}^n$ has one global modulus tending to zero. For example, $f(t)=t^2$ is continuous but not uniformly continuous on $\mathbb{R}$; local compact neighborhoods, not a global bound, are what the pointwise theorem requires.

A Common Pitfall: Signed Averages Can Cancel

The absolute value in the Lebesgue-point definition is essential. For $f(y)=y$ on $\mathbb{R}$ at $x=0$, the incorrect calculation would use the signed average $\frac{1}{2r}\int_{-r}^r(y-0)\,dy=0$ and treat this cancellation as evidence that the values are close to zero on average. But positive and negative values cancel in that integral. The required absolute average is instead

\frac{1}{2r}\int_{-r}^r|y|\,dy =\frac{1}{2r}\left(2\int_0^r y\,dy\right) =\frac{r}{2}.

It is this nonnegative quantity that measures oscillation; here it tends to zero, so the conclusion remains true for the correct reason. As a contrast, continuity cannot simply be omitted from the every-point claim. For $f=\chi_{[0,\infty)}$ on $\mathbb{R}$, $f(0)=1$, and for every $r>0$ the integrand $|f(y)-f(0)|$ equals $1$ on $(-r,0)$ and $0$ on $[0,r)$. Its average over $(-r,r)$ is $r/(2r)=1/2$, not a quantity tending to zero. The function is discontinuous at the point in question.

Check Your Understanding

Use the absolute-average definition and the compact-neighborhood estimate in your reasoning.

For $f(t)=|t|$ on $\mathbb{R}$, compute the average oscillation at an arbitrary center $a$, or give a direct bound that proves $a$ is a Lebesgue point.
Explain why continuity on a compact neighborhood implies local integrability and why that fact makes the displayed averages finite.
For $f(t)=t$ at $0$, compute both the signed average and the average of the absolute difference. Which one appears in the definition?
State how the compact-set corollary controls the averages when the center varies over $K$, and identify which part of its bound tends to zero.
For $f=\chi_{[0,\infty)}$ at $0$, calculate the average oscillation for every $r>0$ and explain why the continuous-function theorem does not apply.

Continuous Functions Have Lebesgue Points

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