Differentiation with Cubes and Balls

Two Families of Neighborhoods

Lebesgue differentiation is often stated using balls, but many arguments and applications use cubes. The two statements are compatible because a centered cube and a centered ball contain one another after a fixed change of scale. The resulting volume ratios depend on the dimension, but not on the location or the size of the neighborhoods. That uniformity is precisely what allows a limit over balls to control a limit over cubes.

Throughout, the measure space is $(\mathbb{R}^n,\mathcal{L},m)$, with complete Lebesgue measure, which is sigma-finite. The functions $f:\mathbb{R}^n\to\mathbb{R}$ or $f:\mathbb{R}^n\to\mathbb{C}$ are Lebesgue measurable and locally integrable: $\int_K |f|\,dm<\infty$ for every compact set $K$. In particular, $f(x)$ is finite for $m$-almost every $x$. All pointwise limits below are asserted at points where $f(x)$ is finite.

Definition: For $x\in\mathbb{R}^n$ and $r>0$, let $B(x,r)=\{y:|y-x|<r\}$ be the open Euclidean ball and let $Q(x,r)=\prod_{j=1}^n(x_j-r,x_j+r)$ be the open axis-parallel cube. Write $\omega_n=m(B(0,1))$, so $m(B(x,r))=\omega_n r^n$, while $m(Q(x,r))=(2r)^n$. For a measurable set $A$ of finite positive measure, the average of $f$ over $A$ is $f_A=m(A)^{-1}\int_A f\,dm$.

The form of differentiation we need concerns the average oscillation from the value at the center, rather than only the signed or complex average. The Lebesgue differentiation theorem on $\mathbb{R}^n$, established for balls earlier, says that for $f\in L^1_{\mathrm{loc}}(\mathbb{R}^n)$,

\lim_{s\downarrow0}\frac{1}{m(B(x,s))} \int_{B(x,s)}|f(y)-f(x)|\,dm(y)=0 \quad\text{for }m\text{-almost every }x.

We will use this ball theorem as the input, and prove the corresponding assertion for cubes. The absolute value is the real absolute value or complex modulus, as appropriate. Once the oscillation average tends to zero, the ordinary average converges too, since the absolute value of an average is bounded by the average of the absolute value.

The Geometric Comparison

Theorem: For every $f\in L^1_{\mathrm{loc}}(\mathbb{R}^n)$, real-valued or complex-valued, the ball and cube oscillation limits are equivalent at every $x$ where $f(x)$ is finite: $$ \lim_{r\downarrow0}\frac{1}{m(Q(x,r))}\int_{Q(x,r)}|f(y)-f(x)|\,dm(y)=0 $$ if and only if $$ \lim_{r\downarrow0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y)=0. $$ Consequently, the ball version of the Lebesgue differentiation theorem implies the cube version for $m$-almost every $x\in\mathbb{R}^n$.

Proof. Fix $x$ with $f(x)$ finite, and set $g_x(y)=|f(y)-f(x)|$. This is a nonnegative measurable function and is integrable on every bounded set: on a bounded set $K$, $\int_K g_x\,dm\le\int_K|f|\,dm+|f(x)|m(K)<\infty$. The geometric containments, valid for every $r>0$, are $B(x,r)\subseteq Q(x,r)\subseteq B(x,\sqrt{n}\,r)$. The first holds because each coordinate of $y-x$ has magnitude at most $|y-x|<r$. The second holds because $y\in Q(x,r)$ gives $|y-x|^2<nr^2$.

For the cube-to-ball comparison, nonnegativity of $g_x$ and $Q(x,r)\subseteq B(x,\sqrt n\,r)$ give $\int_{Q(x,r)}g_x\,dm\le\int_{B(x,\sqrt n\,r)}g_x\,dm$. Dividing by $m(Q(x,r))$ and inserting the ball volume yields

\frac{1}{m(Q(x,r))}\int_{Q(x,r)}g_x\,dm \le \frac{\omega_n n^{n/2}}{2^n} \frac{1}{m(B(x,\sqrt n\,r))} \int_{B(x,\sqrt n\,r)}g_x\,dm.

The multiplier is fixed and finite, and $\sqrt n\,r\downarrow0$ as $r\downarrow0$. Thus, if the ball oscillation average tends to zero, so does the cube oscillation average. This proves one implication without interchanging any limits and integrals.

For the reverse implication, use $B(x,r)\subseteq Q(x,r)$ and the same nonnegativity to obtain

\frac{1}{m(B(x,r))}\int_{B(x,r)}g_x\,dm \le \frac{2^n}{\omega_n} \frac{1}{m(Q(x,r))}\int_{Q(x,r)}g_x\,dm.

Here too the multiplier is fixed and finite. A zero limit for cube oscillations therefore forces a zero limit for ball oscillations. The two limits are equivalent at the chosen $x$. Finally, the ball differentiation theorem gives the ball limit for almost every $x$; the equivalence gives the cube limit at every such point. $\square$

At any point where the oscillation limit is zero, the average itself converges to the point value. Indeed, for either neighborhood $A$,

|f_A-f(x)| =\left|\frac{1}{m(A)}\int_A(f(y)-f(x))\,dm(y)\right| \le\frac{1}{m(A)}\int_A|f(y)-f(x)|\,dm(y).

This inequality applies to real and complex functions and uses only the integral triangle inequality. The theorem therefore also gives $f_{Q(x,r)}\to f(x)$ for $m$-almost every $x$, with the averages defined for sufficiently small $r$.

Worked Comparisons

Worked Example: A Linear Function in the Plane

Let $f(y_1,y_2)=y_1$ on $\mathbb{R}^2$, and fix $x=(a,b)$. Compute its averages over $Q(x,r)$ and $B(x,r)$, and compute both oscillation averages.

Write $y=x+u$. Each neighborhood centered at $x$ is symmetric under $u\mapsto-u$, so the average of $u_1$ is zero. Consequently, both averages of $f(x+u)=a+u_1$ equal $a=f(x)$. The cube oscillation is the average of $|u_1|$. Since the cube has area $4r^2$,

\frac{1}{4r^2}\int_{-r}^{r}\int_{-r}^{r}|u_1|\,du_2\,du_1 =\frac{1}{4r^2}(2r)\left(2\int_0^r u_1\,du_1\right) =\frac{1}{4r^2}(2r)(r^2) =\frac r2.

For the ball, vertical cross-sections at first coordinate $u_1$ have length $2\sqrt{r^2-u_1^2}$. Therefore its oscillation integral is

\int_{B(0,r)}|u_1|\,du =4\int_0^r u_1\sqrt{r^2-u_1^2}\,du_1 =\frac{4r^3}{3}.

Dividing by the disk area $\pi r^2$ gives ball oscillation average $4r/(3\pi)$. Both $r/2$ and $4r/(3\pi)$ tend to zero as $r\downarrow0$, in agreement with differentiation. Check: the signed averages are $a$ by symmetry, and the two computed nonnegative oscillations are positive for $r>0$ and vanish in the limit.

Worked Example: A Quadratic Function Has Different Averages

In $\mathbb{R}^2$, take $f(y)=|y|^2$ and center both neighborhoods at the origin. Here $f(0)=0$, so the oscillation averages are exactly the averages of $f$. For a disk of radius $r$, polar coordinates give

\frac{1}{\pi r^2}\int_0^{2\pi}\int_0^r \rho^2\,\rho\,d\rho\,d\theta =\frac{1}{\pi r^2}\,(2\pi)\frac{r^4}{4} =\frac{r^2}{2}.

For the square $(-r,r)^2$, integrate the two coordinate squares separately. The integral of $y_1^2$ is $(2r)(2r^3/3)=4r^4/3$, and the integral of $y_2^2$ is also $(2r)(2r^3/3)=4r^4/3$. Their sum is $8r^4/3$; dividing by the square area $4r^2$ gives cube average $2r^2/3$. Thus the ball and cube averages are generally unequal, but both tend to $f(0)=0$.

Check: at $r=1$, substitution gives ball average $1^2/2=1/2$ and cube average $2(1^2)/3=2/3$. Their difference is $2/3-1/2=1/6$; at general $r$, the difference is $2r^2/3-r^2/2=r^2/6$, which tends to zero.

Worked Example: Differentiation Can Fail at a Boundary Point

In $\mathbb{R}^2$, let $f(y)=\chi_{\{y_1>0\}}(y)$, and consider $x=0$. The strict inequality in the defining set means $f(0)=0$. In the centered square, the part with $y_1>0$ has area $(r)(2r)=2r^2$, while the whole square has area $4r^2$; its average is therefore $2r^2/(4r^2)=1/2$. In the centered disk, reflection across the line $y_1=0$ divides the disk into two equal halves, and the line itself has two-dimensional measure zero. Its average is also $1/2$.

Neither average tends to $f(0)=0$. This is not a contradiction: differentiation is an almost-everywhere statement, not a statement at every point. Check: for every $r>0$, both averages equal $1/2$, so their limits are $1/2$, not $0$. The example also shows why changing a strict set-membership convention at the single boundary point does not alter either integral.

The Volume Factor and a Common Pitfall

Containment alone does not imply that the average over a smaller set is bounded by the average over a larger set. The numerator of a nonnegative integral increases with the set, but the denominator increases too. The comparison proof must retain the ratio of volumes; discarding it can produce a false inequality.

Worked Example: Why the Volume Ratio Is Necessary

In $\mathbb{R}^2$, fix $r>0$, let $Q=Q(0,r)$, and take $g=\chi_Q$. Since $g=1$ throughout $Q$, its average over $Q$ is $m(Q)/m(Q)=1$. The containing ball $B(0,\sqrt2\,r)$ has area $\pi(\sqrt2\,r)^2=2\pi r^2$. Its intersection with $Q$ is $Q$, up to boundary sets of measure zero, so the ball average is $4r^2/(2\pi r^2)=2/\pi$.

A tempting but incorrect calculation is “$Q\subseteq B$, so the average over $Q$ is at most the average over $B$,” which would assert $1\le2/\pi$. This is false: $2/\pi<1$. The correct comparison includes the volume factor. In dimension two, the factor from the theorem is $\omega_2\,2^{2/2}/2^2=\pi/2$; multiplying the ball average by it gives $(\pi/2)(2/\pi)=1$, exactly the cube average.

Check: the integral over $Q$ is its area $4r^2$, and the integral over the containing ball is also $4r^2$, since $g$ vanishes outside $Q$. Dividing these same integrals by their different areas gives $1$ and $2/\pi$, respectively. The failed inequality is precisely the one that omitted this denominator change.

The key point is that the containment scales are fixed: a cube of radius $r$ fits in a ball of radius $\sqrt n\,r$, while the ball of radius $r$ fits in the cube of radius $r$. The associated volume ratios are finite constants depending only on $n$. Therefore, at any point where one family’s oscillation averages tend to zero, the other family’s do as well. Combined with the ball differentiation theorem, this establishes differentiation by centered cubes for locally integrable functions, almost everywhere with respect to Lebesgue measure.

Check Your Understanding

Use the containments and volume ratios above to organize your reasoning.

Show that $Q(x,r)\subseteq B(x,\sqrt n\,r)$, and compute the ratio $m(B(x,\sqrt n\,r))/m(Q(x,r))$.
Explain why $B(x,r)\subseteq Q(x,r)$ gives a bound for the ball oscillation average by the cube oscillation average with factor $2^n/\omega_n$.
For $f(y)=|y|^2$ in $\mathbb{R}^2$, use the displayed formulas to find the ball and cube averages at the origin when $r=2$.
For the indicator of $\{y_1>0\}$, why does failure of differentiation at the origin not contradict an almost-everywhere theorem?
State precisely why containment of sets by itself does not order their averages.