Ball Averages Recover a Function Almost Everywhere
For a locally integrable function, averages over small balls need not recover its value at every point. They do recover it at almost every point, even when the function is unbounded and the domain has infinite measure. The proof combines two ideas: continuous functions have vanishing local oscillation, and the maximal-function estimate controls the points where an \(L^1\) approximation has a large local average.
Throughout, \(m\) denotes Lebesgue measure on \(\mathbb{R}^n\), and \(B(x,r)=\{y\in\mathbb{R}^n:|y-x|<r\}\). Write \(\omega_n=m(B(0,1))\), so \(m(B(x,r))=\omega_n r^n\). A real-valued or complex-valued measurable function \(f\) is locally integrable if \(\int_K |f|\,dm<\infty\) for every compact set \(K\subset\mathbb{R}^n\). Such a function is finite almost everywhere; choose a measurable representative that is finite everywhere by redefining it on the null set where it is infinite. The theorem concerns this representative’s point values and asserts the conclusion almost everywhere.
This definition uses the average of the absolute difference, not the absolute difference of averages. That distinction measures the local variation around the particular value \(f(x)\).
Two Ingredients from the Earlier Material
We use continuous approximation in \(L^1\): if \(h\in L^1(\mathbb{R}^n)\) and \(\eta>0\), there is a continuous, integrable function \(g\) such that \(\|h-g\|_1=\int_{\mathbb{R}^n}|h-g|\,dm<\eta\). We also use the weak-type estimate for the centered Hardy–Littlewood maximal function. For \(h\in L^1(\mathbb{R}^n)\), define $$ Mh(x)=\sup_{r>0}\frac{1}{\omega_n r^n}\int_{B(x,r)}|h(y)|\,dm(y). $$ The estimate states that there is a constant \(C_n<\infty\), depending only on \(n\), such that for every \(\lambda>0\), $$ m(\{x:Mh(x)>\lambda\})\le \frac{C_n}{\lambda}\|h\|_1. $$ Both results apply to Lebesgue measure on \(\mathbb{R}^n\); the approximation result concerns integrable functions, and the maximal estimate requires \(h\in L^1\) and a strictly positive threshold. These are the continuous-approximation theorem and the weak-type maximal estimate established in the earlier material.
Proof. Let $$ D(x)=\limsup_{r\downarrow0} A_r f(x). $$ The averages are nonnegative, so \(D(x)\in[0,\infty]\). The function \(D\) is measurable: for fixed \(r>0\), \(x\mapsto\int_{B(x,r)}|f(y)-f(x)|\,dm(y)\) is measurable by measurability of the integrand in \((x,y)\) and the theorem on integration of measurable sections (Tonelli’s theorem, applied locally to nonnegative functions). For each fixed \(x\), this integral is continuous as a function of \(r>0\). Indeed, on bounded ranges of radii the integrand is integrable on a bounded region, and the indicators of the balls converge almost everywhere as the radius varies; the boundary sphere has Lebesgue measure zero. The denominator is also continuous and positive for \(r>0\). Consequently, the limsup as \(r\downarrow0\) can be taken over positive rational radii, which makes \(D\) measurable.
Fix an integer \(R\ge1\) and set \(U=B(0,R)\). Define \(f_R=f\chi_{B(0,R+1)}\). Local integrability implies \(f_R\in L^1(\mathbb{R}^n)\). For \(x\in U\) and \(0<r<1\), \(B(x,r)\subset B(0,R+1)\). Thus \(f_R=f\) throughout that ball and at \(x\), so the averages defining \(D(x)\) can be computed using \(f_R\).
For any \(\eta>0\), choose a continuous \(g\in L^1(\mathbb{R}^n)\) with \(\|f_R-g\|_1<\eta\), and put \(h=f_R-g\). For \(x\in U\), the triangle inequality gives, for \(0<r<1\), $$ A_r f(x)\le \frac{1}{\omega_n r^n}\int_{B(x,r)}|h(y)|\,dm(y) +|h(x)| +\frac{1}{\omega_n r^n}\int_{B(x,r)}|g(y)-g(x)|\,dm(y). $$ The first term is at most \(Mh(x)\). The final term tends to zero as \(r\downarrow0\): continuity of \(g\) at \(x\) means that for every \(\varepsilon>0\), \(|g(y)-g(x)|<\varepsilon\) whenever \(|y-x|\) is sufficiently small, so the average is then at most \(\varepsilon\). Therefore $$ D(x)\le Mh(x)+|h(x)| \qquad (x\in U). $$
Fix \(a>0\). If \(x\in U\) and \(D(x)>a\), then at least one of \(Mh(x)>a/2\) or \(|h(x)|>a/2\) holds. Otherwise their sum would be at most \(a\), contradicting \(D(x)>a\). The weak-type estimate and Chebyshev’s inequality therefore yield $$ m(\{x\in U:D(x)>a\}) \le \frac{2C_n}{a}\|h\|_1+\frac{2}{a}\|h\|_1 <\frac{2(C_n+1)}{a}\eta. $$ This holds for every \(\eta>0\), choosing a new approximation \(g\) when necessary. The left side is independent of \(\eta\), so it must be zero. Taking \(a=1/k\) for each positive integer \(k\) shows that \(\{x\in U:D(x)>0\}\), the union of the sets \(\{x\in U:D(x)>1/k\}\), is null. Since \(\mathbb{R}^n=\bigcup_{R\ge1}B(0,R)\), the set where \(D(x)>0\) is null in all of \(\mathbb{R}^n\). Nonnegativity of \(A_r f(x)\) now implies \(A_r f(x)\to0\) outside that null set. \(\square\)
Worked Examples
Worked Example: A Continuous Function on the Line
Take \(n=1\), Lebesgue measure \(m\) on \(\mathbb{R}\), and \(f(x)=x\). Compute the mean absolute oscillation at any \(x\in\mathbb{R}\).
Here \(B(x,r)=(x-r,x+r)\), whose measure is \(2r\). Substitute \(y=x+t\): $$ A_r f(x)=\frac{1}{2r}\int_{-r}^{r}|(x+t)-x|\,dt =\frac{1}{2r}\int_{-r}^{r}|t|\,dt =\frac{1}{2r}\left(2\int_0^r t\,dt\right) =\frac{1}{2r}r^2=\frac r2. $$ Thus \(A_r f(x)\to0\) as \(r\downarrow0\), for every \(x\). Check: for \(r=2\), the average is \(2/2=1\); directly, \(\int_{x-2}^{x+2}|y-x|\,dy=2\int_0^2t\,dt=4\), and \(4/(2\cdot2)=1\).
Worked Example: A Step Function and Its Exceptional Point
On \(\mathbb{R}\), let \(f=\chi_{[0,\infty)}\). For \(x>0\), if \(0<r<x\), then every \(y\in(x-r,x+r)\) is positive and \(f(y)=f(x)=1\), so \(A_r f(x)=0\). For \(x<0\), if \(0<r<-x\), every \(y\) in the ball is negative and \(f(y)=f(x)=0\), again giving \(A_r f(x)=0\).
At \(x=0\), \(f(0)=1\). For every \(r>0\), the integrand \(|f(y)-f(0)|\) equals \(1\) on \((-r,0)\) and \(0\) on \([0,r)\). Hence $$ A_r f(0)=\frac{1}{2r}\int_{-r}^{0}1\,dy=\frac{r}{2r}=\frac12. $$ The limit at zero is not zero, but the exceptional set \(\{0\}\) has Lebesgue measure zero. Check: when \(r=3\), the integral is \(3\), the ball has measure \(6\), and the average is \(3/6=1/2\).
Worked Example: Changing Values on a Null Set
Let \(f=\chi_{\mathbb{Q}}\) on \(\mathbb{R}\). The rationals are a countable, hence Lebesgue-null, set. If \(x\notin\mathbb{Q}\), then \(f(x)=0\), and \(f(y)=1\) only on the null set \(\mathbb{Q}\). Therefore, for every \(r>0\), $$ A_r f(x)=\frac{1}{2r}\int_{x-r}^{x+r}\chi_{\mathbb{Q}}(y)\,dy=0. $$ If \(x\in\mathbb{Q}\), then \(f(x)=1\), and \(|f(y)-f(x)|=1\) at every irrational \(y\). The rationals in the interval have measure zero, so the integral is \(2r\) and \(A_r f(x)=2r/(2r)=1\). Thus the conclusion holds at every irrational and fails at every rational. The failure set is null, as the theorem permits. This function is locally integrable, with integral zero on every bounded interval, despite its values at rational points.
Check: for \(r=1\), the average at a rational center is \(2/2=1\); at an irrational center it is \(0/2=0\). The two values differ because the chosen pointwise representative differs on \(\mathbb{Q}\), while the almost-everywhere conclusion is unchanged.
A Common Pitfall and Why the Absolute Value Matters
A tempting but invalid shortcut is to replace the average of the absolute difference by the absolute difference between the average and the point value. In general, the triangle inequality gives $$ \left|\frac{1}{m(B(x,r))}\int_{B(x,r)}f(y)\,dm(y)-f(x)\right| \le A_r f(x), $$ not equality. Cancellation can make the left side zero even when the right side is positive. For \(f(y)=y\) on \(\mathbb{R}\), centered at \(x=0\), symmetry gives \(\frac{1}{2r}\int_{-r}^{r}y\,dy=0\), so the absolute difference of averages is \(0\). But the correct mean absolute oscillation is \(r/2\), as computed above. The correct quantity still tends to zero, but the cancellation calculation alone does not establish that fact.
The result is useful because it identifies local averages with the function’s pointwise values at almost every point, even when \(f\) is only locally integrable. It also explains why a theorem about almost every point is the natural conclusion: changing a measurable function on a null set can change the averages centered at points of that set, as the rational-indicator example shows. The theorem does not promise convergence at every point, nor does it allow the absolute value inside the average to be discarded.
Check Your Understanding
In each question, distinguish the pointwise claim from the almost-everywhere conclusion.
- For \(f(x)=|x|\) on \(\mathbb{R}\), compute \(A_r f(0)\) and determine its limit as \(r\downarrow0\).
- For \(f=\chi_{[0,\infty)}\), calculate \(A_r f(0)\) and explain why this does not contradict the differentiation theorem.
- Why does local integrability suffice in the proof even though the weak-type maximal estimate is applied to an \(L^1(\mathbb{R}^n)\) function?
- For \(f=\chi_{\mathbb{Q}}\), at which points does the mean absolute oscillation tend to zero, and what is the measure of the set where it does not?
- State the correct inequality relating the absolute difference of the average of \(f\) and \(f(x)\) to the mean absolute oscillation. Why does the reverse equality generally fail?