Theorem 2.8.3 (Lebesgue differentiation)CF98893DE93E

Proof. We prove a maximal inequality on $\mathbb R^ d$, deduce the theorem for integrable functions by approximation with continuous ones, and then localize. The set $N$ where $f$ is infinite is a null set. Changing $f$ to $0$ on $N$ changes neither the averages in the statement nor the values $f(x)$ for $x\notin N$, so the modified function has the same Lebesgue points outside $N$ and the same mollifications. We may therefore assume that $f$ is finite everywhere.

Consider first $f\in L^1(\mathbb R^ d)$, and define its maximal function $Mf(x)$ as the supremum over $r>0$ of the averages of $|f|$ over $Q(x,r)$. For fixed $r$, the average of $|f|$ over $Q(x,r)$ depends continuously on $x$, because $\int _{Q(x,r)}|f|$ and $\int _{Q(x',r)}|f|$ differ by at most $\| f(\cdot +x-x')-f\| _1$, which tends to zero as $x'\to x$ by the continuity of translation in $L^1$ (lemma 2.5.1(2)). As a supremum of continuous functions, $Mf$ is lower semicontinuous, so $\{ Mf>t\} $ is open. We claim the maximal inequality

By inner regularity of Lebesgue measure, it suffices to bound $|L|$ for compact $L\subset \{ Mf>t\} $. Each point of $L$ is the center of an open cube over which the average of $|f|$ exceeds $t$, and finitely many of these cubes cover $L$. Select one of largest side length, discard all cubes meeting it, and repeat with the remaining ones. The selected cubes are disjoint, and each discarded cube meets a selected cube of at least the same side length, so it lies in the concentric cube with three times that side length. These enlarged cubes cover $L$, and therefore $|L|$ is at most $3^ d$ times the total volume of the selected cubes, which is less than $t^{-1}\| f\| _1$.

Now let $g\in C_ c(\mathbb R^ d)$. For every $x$ and $r$,

and the last term tends to zero as $r\downarrow 0$ because $g$ is continuous at $x$. Hence the upper limit $\Lambda f(x)$ of the left side satisfies $\Lambda f\leq M(f-g)+|f-g|$, and for $t>0$ the set $\{ \Lambda f>2t\} $ is contained in $\{ M(f-g)>t\} \cup \{ |f-g|>t\} $, whose measure is at most $(3^ d+1)t^{-1}\| f-g\| _1$ by the maximal inequality and Chebyshev's inequality. By proposition 2.3.5, $\| f-g\| _1$ can be made arbitrarily small, so $\{ \Lambda f>2t\} $ is contained in measurable sets of arbitrarily small measure and is therefore a null set. Letting $t$ run through the positive rationals shows that $\Lambda f=0$ almost everywhere.

Next, let $f$ be locally integrable on $\Omega $, and let $(K_ j)$ be a compact exhaustion of $\Omega $. The function equal to $f$ on $\operatorname{Int}K_{j+1}$ and to $0$ elsewhere belongs to $L^1(\mathbb R^ d)$, and for $x\in \operatorname{Int}K_ j$ and small $r$ the averages in the statement are the same for $f$ and for this function. Hence almost every point of each $\operatorname{Int}K_ j$, and therefore of $\Omega $, is a Lebesgue point of $f$.

Finally, let $x$ be a Lebesgue point of $f$ and let $\varepsilon >0$ be so small that $x\in \Omega _\varepsilon $. The function $\rho _\varepsilon (x-\cdot )$ is bounded by $\varepsilon ^{-d}\sup \rho =2^ d\sup \rho /|Q(x,\varepsilon )|$ and vanishes outside $B(x,\varepsilon )\subset Q(x,\varepsilon )$. Since $\int \rho _\varepsilon =1$,

which tends to zero as $\varepsilon \to 0$. $\square$