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Differentiation of L-one functions for a doubling weight

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain), hence Countable Choice (Dependent choice implies countable choice). Let v be a weight on Rn whose measure v dλ is doubling, and let h∈Lloc1(v) (Weights, their associated measures, and the spaces L^p(w)). Then for v-almost every x∈Rn, lim⁡r→0+1v(B(x,r))∫B(x,r)h v dλ=h(x); more generally the same limit holds along any family of balls or cubes shrinking nicely to x (A family shrinking nicely to a point) with v-measure comparable to the corresponding ball.

Facts & Assumptions

Given: Dependent Choice, a weight v with v dλ doubling, and h∈Lloc1(v).

[F1]

Mvh(x)=sup⁡r>0v(B(x,r))−1∫B(x,r)∣h∣v dλ is the weighted maximal function of The weighted maximal function of a doubling weight, and it obeys the weak (1,1) bound v({Mvg>η})≤C(n,cv)η−1∫∣g∣v dλ for every g∈L1(v) (The weighted maximal function of a doubling weight is weak (1,1)).

[F2]

Cc(Rn) is dense in L1(v dλ) because v dλ is a Radon measure (C_c(X) is dense in L^p(mu) for a Radon measure), with complex density obtained by approximating the two real components separately; a function g∈Cc is continuous, so for every x and every family shrinking nicely to x the weighted averages of g converge to g(x) (the averages of ∣g(⋅)−g(x)∣ are bounded by the maximum of ∣g−g(x)∣ over the shrinking sets, which tends to 0).

[F3]

Chebyshev's inequality: for a nonnegative measurable G and η>0, v({G>η})≤η−1∫G v dλ (Chebyshev-Markov inequality for the integral).

[F4]

Dominated convergence gives continuity in radius for local weighted integrals, and Lebesgue-measurable functions have Borel representatives: apply A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra componentwise using L(Rn) is exactly the completion of the restriction of λn to the Borel sets, setting infinite values on null sets to zero. These representatives give jointly Borel integrands by composition with continuous maps; Tonelli gives measurable section integrals (Dominated convergence, Borel representatives make the convolution integrand Borel measurable, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

Proof

technique · direct
1.1F1F2F4givenalgebra

First suppose h∈L1(v) globally and put μ=v dλ. For g∈Cc and G=∣h−g∣, the centred weighted oscillation Dh(x):=lim sup⁡r↓0μ(B(x,r))−1∫B(x,r)∣h(y)−h(x)∣ dμ(y) is at most MvG(x)+G(x), since the corresponding oscillation of continuous g tends to zero. These limsups are measurable: for each fixed x the integrals are continuous in positive radius by dominated convergence, hence rational radii suffice; joint measurability follows after choosing Borel representatives, and changing them on a null set does not change the a.e. assertion.

2.1F1F2F3step 1.1givenalgebra

For η>0, {Dh>η}⊆{MvG>η/2}∪{G>η/2}. The weak bound and Chebyshev give μ({Dh>η})≤2(C(n,cv)+1)η−1∥h−g∥L1(v). Taking the infimum over g∈Cc using [F2] makes this measure zero. The countable union over η=1/k is null, so the centred absolute oscillation tends to zero a.e. For local h, apply this result to hm=h1B(0,m+1)∈L1(v); on B(0,m), sufficiently small centred balls see h=hm. The countable union of these exceptional sets is null and ⋃mB(0,m)=Rn, proving the same conclusion for every local input.

3.1step 2.1givenalgebra∎

Outside the null set of step 2.1, the absolute oscillation over any shrinking set Er⊆B(x,ρr) with ρr→0 and μ(B(x,ρr))≤Cxμ(Er) is bounded by Cx times the centred oscillation. It therefore tends to zero, and the modulus of the difference between the average of h and h(x) is at most this oscillation. Balls and cubes shrinking nicely with the stipulated weighted comparability meet these hypotheses. This proves all the stated limits, on a common full-measure set.

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