Ball Averages and Lebesgue Points

Local Averages and Pointwise Behavior

For an integrable function, the value at one point does not affect its integral. Yet the value at a point is part of the question when we ask whether nearby values settle around that particular value. Ball averages make this distinction precise: they summarize behavior on shrinking neighborhoods, while the average absolute deviation tests whether those nearby values are close to the value assigned at the center.

Throughout, \(m\) is Lebesgue measure on \(\mathbb{R}^n\), and \(f:\mathbb{R}^n\to\mathbb{R}\) is a measurable, finite-valued function that is locally integrable: \(\int_K |f|\,dm<\infty\) for every compact set \(K\subset\mathbb{R}^n\). We use open balls \(B(x,r)=\{y\in\mathbb{R}^n:|y-x|<r\}\), where \(r>0\). Their measure is \(m(B(x,r))=\omega_n r^n\), with \(\omega_n=m(B(0,1))>0\), so every average below is normalized by a finite, positive measure.

Definition: The ball average of \(f\) at \(x\) with radius \(r\) is \(\displaystyle A_r f(x)=\frac{1}{m(B(x,r))}\int_{B(x,r)} f(y)\,dm(y)\). The average absolute deviation from the value at the center is \(\displaystyle D_r f(x)=\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y)\). The upper average oscillation at \(x\) is \(\displaystyle \overline{\omega}(f,x)=\limsup_{r\downarrow0}D_r f(x)\), with values in \([0,\infty]\). The point \(x\) is a Lebesgue point of \(f\) if \(\overline{\omega}(f,x)=0\), equivalently if \(D_r f(x)\to0\) as \(r\downarrow0\).

The average \(A_r f(x)\) and the deviation \(D_r f(x)\) answer different questions. In particular, \(D_r f(x)\) measures closeness to the specified value \(f(x)\), not closeness to the average \(A_r f(x)\). The limit or upper limit uses all positive radii tending to zero; it is not enough that the desired behavior hold for just one sequence of radii.

Continuous Functions

Proposition: If \(f:\mathbb{R}^n\to\mathbb{R}\) is continuous at \(x\), then \(x\) is a Lebesgue point of \(f\), and \(A_r f(x)\to f(x)\) as \(r\downarrow0\).

Proof. Fix \(x\) at which \(f\) is continuous, and let \(\varepsilon>0\). By continuity there is a \(\delta>0\) such that \(|f(y)-f(x)|<\varepsilon\) whenever \(|y-x|<\delta\). For every \(r\) with \(0<r<\delta\), this inequality holds at every \(y\in B(x,r)\). Integrating it and dividing by the positive measure \(m(B(x,r))\) gives \[ 0\le D_r f(x) =\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y) \le\varepsilon. \] The same bound on the difference of averages follows from the integral triangle inequality: \[ |A_r f(x)-f(x)| =\left|\frac{1}{m(B(x,r))}\int_{B(x,r)}(f(y)-f(x))\,dm(y)\right| \le D_r f(x)\le\varepsilon. \] Thus both quantities tend to zero, so \(\overline{\omega}(f,x)=0\) and \(A_r f(x)\to f(x)\). \(\square\)

Worked Example: A Linear Function on the Real Line

Let \(f(t)=t\) on \(\mathbb{R}\), with Lebesgue measure \(m\). The ball \(B(x,r)\) is the interval \((x-r,x+r)\), of length \(2r\). Direct integration gives \[ A_r f(x)=\frac{1}{2r}\int_{x-r}^{x+r}t\,dt =\frac{(x+r)^2-(x-r)^2}{4r} =\frac{4xr}{4r}=x. \] For the deviation, substitute \(u=t-x\). Then \[ D_r f(x)=\frac{1}{2r}\int_{x-r}^{x+r}|t-x|\,dt =\frac{1}{2r}\int_{-r}^{r}|u|\,du =\frac{1}{2r}\left(2\int_0^r u\,du\right) =\frac{r^2}{2r}=\frac r2. \] Consequently \(\overline{\omega}(f,x)=\lim_{r\downarrow0}r/2=0\) at every \(x\). Check: the average is \(x\), as symmetry about \(x\) predicts, and the deviation is positive for each \(r>0\) but tends to zero.

Indicators and Density

For a Lebesgue measurable set \(E\subset\mathbb{R}^n\), its indicator \(\chi_E\) equals \(1\) on \(E\) and \(0\) on \(E^c\). It is locally integrable because it is bounded by \(1\) and every ball has finite measure. Its ball average is a proportion: \[ A_r\chi_E(x)=\frac{m(E\cap B(x,r))}{m(B(x,r))}. \] This ratio lies in \([0,1]\). It is the fraction of the ball occupied by \(E\), with membership understood exactly as specified by the set \(E\).

Proposition: For a Lebesgue measurable \(E\subset\mathbb{R}^n\), the upper average oscillation of \(\chi_E\) at \(x\) is \(\displaystyle\overline{\omega}(\chi_E,x)=\limsup_{r\downarrow0}\left|A_r\chi_E(x)-\chi_E(x)\right|\). In particular, if \(x\in E\), then \(x\) is a Lebesgue point of \(\chi_E\) exactly when \(A_r\chi_E(x)\to1\); if \(x\notin E\), it is a Lebesgue point exactly when \(A_r\chi_E(x)\to0\).

Proof. If \(x\in E\), then \(\chi_E(x)=1\). On \(E\cap B(x,r)\), the value \(|\chi_E(y)-1|\) is zero; on \(E^c\cap B(x,r)\), it is one. Therefore \[ D_r\chi_E(x)=\frac{m(E^c\cap B(x,r))}{m(B(x,r))} =1-\frac{m(E\cap B(x,r))}{m(B(x,r))} =1-A_r\chi_E(x) =|A_r\chi_E(x)-\chi_E(x)|. \] If \(x\notin E\), then \(\chi_E(x)=0\). The absolute difference is one on \(E\cap B(x,r)\) and zero on its complement in the ball, so \[ D_r\chi_E(x)=\frac{m(E\cap B(x,r))}{m(B(x,r))} =A_r\chi_E(x) =|A_r\chi_E(x)-\chi_E(x)|. \] These two cases prove the identity for every \(x\). Since the ratios lie in \([0,1]\), the deviation tends to zero exactly when the ratio tends to the value \(\chi_E(x)\), proving both characterizations. \(\square\)

Thus, at a point of \(E\), the Lebesgue-point condition says that the proportion of \(E\) in small balls tends to one. At a point outside \(E\), it says that this proportion tends to zero. These are density conditions, not merely statements about whether the point lies in the topological boundary. The proposition concerns the particular representative \(\chi_E\), including its specified value at \(x\).

Worked Example: The Indicator of a Half-Line at Its Endpoint

Take \(E=(0,\infty)\subset\mathbb{R}\) and \(x=0\). Since \(0\notin E\), \(\chi_E(0)=0\). For every \(r>0\), \(B(0,r)=(-r,r)\), and \(E\cap B(0,r)=(0,r)\). The interval \((0,r)\) has measure \(r\), whereas \((-r,r)\) has measure \(2r\). Hence \[ A_r\chi_E(0)=\frac{r}{2r}=\frac12, \qquad D_r\chi_E(0)=\frac{1}{2r}\int_{-r}^{r}|\chi_E(y)-0|\,dy =\frac{r}{2r}=\frac12. \] Thus \(\overline{\omega}(\chi_E,0)=1/2\), so \(0\) is not a Lebesgue point. Check: the density ratio is \(1/2\) at every radius and does not tend to \(\chi_E(0)=0\).

Worked Example: The Indicator of an Interval

Let \(E=(0,2)\subset\mathbb{R}\). At \(x=1\), if \(0<r<1\), then \(B(1,r)=(1-r,1+r)\subset E\). Therefore \(A_r\chi_E(1)=1\), \(D_r\chi_E(1)=0\), and \(1\) is a Lebesgue point.

At \(x=0\), we have \(0\notin E\). For \(0<r<2\), \(E\cap B(0,r)=(0,r)\), so \(A_r\chi_E(0)=r/(2r)=1/2\), and \(D_r\chi_E(0)=1/2\). The endpoint is not a Lebesgue point. At \(x=3\), if \(0<r<1\), then \(B(3,r)\cap E=\varnothing\), giving \(A_r\chi_E(3)=D_r\chi_E(3)=0\); hence \(3\) is a Lebesgue point. Check: these computations use strict membership in the open interval \((0,2)\), so both endpoints have indicator value zero.

Point Values and a Common Pitfall

A locally integrable function is determined by its values up to changes on null sets for purposes of integration. The definition of Lebesgue point also uses the single value \(f(x)\), however, so changing a function at one point can change whether that point is a Lebesgue point. This does not contradict local integrability: integrals over balls are unchanged by a change on a null set, but the comparison value at the center may change.

Worked Example: Changing a Function at One Point

Define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(0)=1\) and \(g(y)=0\) for \(y\ne0\). This function is measurable and locally integrable; it is zero \(m\)-almost everywhere. For every \(r>0\), the integral of \(g\) over \((-r,r)\) is zero, because the singleton \(\{0\}\) has Lebesgue measure zero. But \(g(0)=1\), and \(|g(y)-g(0)|=1\) for every \(y\ne0\). Consequently \[ A_r g(0)=0,\qquad D_r g(0)=\frac{1}{2r}\int_{-r}^{r}|g(y)-1|\,dy =\frac{2r}{2r}=1. \] Thus \(\overline{\omega}(g,0)=1\), and \(0\) is not a Lebesgue point of this representative. If instead the value at zero is set to \(0\), the resulting function is identically zero, and \(D_r g(0)=0\) for every \(r>0\). The integrals are the same, but the pointwise Lebesgue-point status differs.

Common mistake. It is generally wrong to replace the average absolute deviation by the absolute difference between the average and the center value. For \(f(t)=t\) at \(x=0\), symmetry gives \(A_r f(0)=0=f(0)\), so \(|A_r f(0)-f(0)|=0\). But the direct calculation gives \[ D_r f(0)=\frac{1}{2r}\int_{-r}^{r}|t|\,dt =\frac{1}{2r}\left(2\cdot\frac{r^2}{2}\right) =\frac r2>0. \] The signed deviations cancel in the average; the absolute values do not. The point is still a Lebesgue point because \(r/2\to0\), not because the deviation is zero at each radius.

Ball averages and upper average oscillation provide the local quantities used in differentiation theory. Continuity gives a direct pointwise guarantee, while indicator functions translate the same condition into density ratios. The distinction between integral behavior and the chosen point value remains important whenever the function is specified only up to equality almost everywhere.

Check Your Understanding

For each question, use the normalized ball average and compare the average absolute deviation with the value at the center.

For \(f(t)=t^2\) on \(\mathbb{R}\), compute \(A_r f(x)\) and determine \(\overline{\omega}(f,x)\).
Let \(E=(-1,1)\). Find \(A_r\chi_E(1)\) and \(D_r\chi_E(1)\) for \(0<r<2\). Is \(1\) a Lebesgue point of \(\chi_E\)?
For an arbitrary measurable set \(E\subset\mathbb{R}^n\), state the density-ratio condition for \(x\) to be a Lebesgue point of \(\chi_E\), separately for \(x\in E\) and \(x\notin E\).
Explain why \(|A_r f(x)-f(x)|\) need not equal \(D_r f(x)\), using \(f(t)=t\) at \(x=0\).
Can changing a locally integrable function at one point affect its ball averages? Can it affect its Lebesgue-point status at that point? Justify both answers.