Past exam of the real and harmonic analysis course of the University of Oxford / 2008 / 4

The centered Hardy-Littlewood maximal function isThe Hardy-Littlewood maximal inequality gives, for example,This valid constant follows from the disjoint-interval covering argument: a compact subset of the superlevel set has a finite cover by intervals of average above $t$; select disjoint intervals whose triples cover it, bound their total length by $\Vert f{\Vert }_1/t$, and use inner regularity. An optimal constant is not required.

Let $Df(x)$ denote the upper limit of the displayed mean oscillation as $r\downarrow 0$. For a continuous integrable approximation $g$, continuity gives $Dg(x)=0$. With $h=f-g$, the triangle inequality yieldsTherefore the maximal inequality and Chebyshev's inequality giveDensity makes the right side arbitrarily small. For each positive rational $\epsilon$ the exceptional set is null, proving $Df=0$ almost everywhere, the Lebesgue differentiation theorem. Values where the chosen representative is not finite already form a null set.

For the fundamental theorem define $F(x)={\int }_a^{x}f(t)dt$, with oriented integration for $x<a$. This function is absolutely continuous: for disjoint intervals the sum of $|F(b_j)-F(a_j)|$ is at most the integral of $|f|$ over their union, which is small when its measure is small by absolute continuity of the integral. At a Lebesgue point of $f$, for $h\ne 0$,Thus $F^{\prime }=f$ almost everywhere and $F(b)-F(a)={\int }_a^{b}f$. Differentiability at every point is not claimed for an arbitrary integrable function.

For a set $M$ of finite measure apply the preceding result to $f=1_M$. Since its local average minus its point value is bounded in absolute value by its mean absolute oscillation, it givesFor a general measurable $M$, apply the same argument to $1_{M\cap [-j,j]}\in L^1(\mathbb{R})$. At every interior point of $(-j,j)$, sufficiently short averages agree with those of $1_M$. The countable union of these interiors covers the line, and a countable union of null exceptional sets remains null. This proves the Lebesgue density theorem without a finite-measure assumption on $M$. The source's $\chi$ is the indicator of $M$.

Every $x$ in the maximal superlevel set has a first index $k(x)$ at which the average on its unique cell $M_k(x)\in {\mathcal{P}}_k$ exceeds $t$. Refinement implies that all earlier cells containing $x$ contain this cell, and their averages are at most $t$. Thus $M_{k(x)}(x)\in {\mathcal{S}}_t$ and the selected cells cover the superlevel set.

Two cells at different levels are either disjoint or nested, because of refinement and the partition property. If selected cells of levels $k<l$ intersect, the level-$l$ one lies in the earlier cell whose average is above $t$, contradicting its first-crossing property. Cells at the same level are disjoint unless identical. Hence the selected cells form a countable disjoint partition of the superlevel set. A cell repeated unchanged at successive levels is counted once, at its first crossing; equal earlier ancestors are included in the comparison condition.

The first-crossing maximal inequality for refining partitions follows from countable disjointness and the strict average bound:Thus this partition maximal operator has weak-type constant one.

The conclusion of differentiation along refining metric partitions isThe crucial geometric property is shrinking diameter, not merely refinement: since $x\in M_k(x)$ and $\mathrm{diam}{M}_k(x)<1/k$, continuity makes the corresponding oscillation tend to zero for every continuous test function. Continuous $L^1$ approximation and the maximal inequality then transfer this to $f$, exactly as before: the exceptional-set bound is $2\Vert f-g{\Vert }_1/\epsilon$ for an oscillation threshold $2\epsilon$. Countability makes the maximal function measurable, and positive cell measure makes every average defined. Refinement supplied the disjoint first-crossing argument; shrinking diameter supplies differentiation.