Differentiation of sigma-finite Borel measures finite on compact sets

By [L1], write $\nu ={\nu }_a+{\nu }_s$ with ${\nu }_a=fd\lambda$, ${\nu }_a\ll \lambda$, and ${\nu }_s\perp \lambda$. Choose a Borel set $N$ with $\lambda (N)=0$ on which ${\nu }_s$ is concentrated. For every Borel set $E$, absolute continuity and concentration give ${\nu }_s(E)={\nu }_s(E\cap N)=\nu (E\cap N)\ge 0,{\nu }_a(E)={\nu }_a(E\setminus N)=\nu (E\setminus N)\ge 0.$ Thus both components are positive and ${\nu }_s\le \nu$. Changing $f$ on a null set does not affect the claim, so take $f\ge 0$. Since every closed Euclidean ball is compact, for every $x$ and $R>0$, ${\int }_{B(x,R)}\mathrm{\mid }f\mathrm{\mid }d\lambda ={\nu }_a(B(x,R))\le \nu (\overline{B(x,R)})<\mathrm{\infty }\mathrm{.}$ Hence $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$.

Let $(E_r(x){)}_{r>0}$ be a family of Borel sets shrinking nicely to $x$ with constant ${\alpha }_x>0$. Then $\frac{\nu (E_r(x))}{\lambda (E_r(x))}=\frac{1}{\lambda (E_r(x))}{\int }_{E_r(x)}fd\lambda +\frac{{\nu }_s(E_r(x))}{\lambda (E_r(x))},$ while positivity and the defining comparison give $0\le \frac{{\nu }_s(E_r(x))}{\lambda (E_r(x))}\le \frac{1}{{\alpha }_x}\frac{{\nu }_s(B(x,r))}{\lambda (B(x,r))}\mathrm{.}$ Consequently [L2] and step 1.1 reduce both conclusions to proving $\frac{{\nu }_s(B(x,r))}{\lambda (B(x,r))}⟶0$ for almost every $x$.

Put $G:={\mathbb{R}}^n\setminus N$, so $G$ is Borel, $\lambda ({\mathbb{R}}^n\setminus G)=0$, and ${\nu }_s(G)=0$. For each $m\ge 1$, define $D_m(x):={\sup }_{0<r<1\mathrm{/}m}\frac{{\nu }_s(B(x,r))}{\lambda (B(x,r))}\mathrm{.}$ For fixed $r>0$, if $0<\delta <r$ and $\mathrm{\parallel }x-y{\mathrm{\parallel }}_2<\delta$, then $B(x,r-\delta )\subseteq B(y,r)\mathrm{.}$ By [L5], ${\nu }_s(B(x,r-\delta ))\uparrow {\nu }_s(B(x,r))$ as $\delta \downarrow 0$, so $x\mapsto {\nu }_s(B(x,r))$ is lower semicontinuous. Translation invariance gives the fixed positive denominator $\lambda (B(x,r))=\lambda (B(0,r))$, so each $D_m$ is lower semicontinuous and each set $\{x\in {\mathbb{R}}^n:D_m(x)>\frac{1}{k}\}$ is open. For $k\ge 1$, put $F_k:=G\cap {\bigcap }_{m\ge 1}\{x\in {\mathbb{R}}^n:D_m(x)>\frac{1}{k}\}\mathrm{.}$ Then $F_k$ is Borel. If $x\in G$ and $L(x):={\mathrm{lim\hspace{0.17em}sup}}_{r\to 0^{+}}\frac{{\nu }_s(B(x,r))}{\lambda (B(x,r))}>0,$ choose $k$ with $1\mathrm{/}k<L(x)$. Since $D_m(x)\ge L(x)>1\mathrm{/}k$ for every $m$, one has $x\in F_k$. Thus $\{x\in G:L(x)>0\}\subseteq {\bigcup }_{k\ge 1}{F}_k,$ and it is enough to prove $\lambda (F_k)=0$ for every $k$.

Fix $k\ge 1$ and $\epsilon >0$. Because ${\nu }_s(G)=0$ and ${\nu }_s$ is a Borel measure finite on compact sets, [F1] gives an open set $U_{\epsilon }\supseteq G$ with ${\nu }_s(U_{\epsilon })<\epsilon$. Let $K\subseteq F_k$ be compact, and let $\mathcal{B}$ be the family of all balls $B(x,r)$ such that $x\in K,B(x,r)\subseteq U_{\epsilon },{\nu }_s(B(x,r))>\frac{1}{k}\lambda (B(x,r))\mathrm{.}$ This family covers $K$: for any $x\in K$, openness gives an $m\ge 1$ with $B(x,1\mathrm{/}m)\subseteq U_{\epsilon }$, and $x\in F_k$ gives an $r<1\mathrm{/}m$ satisfying the displayed strict inequality. Compactness supplies a finite subcover of $K$ from $\mathcal{B}$. Apply [L3] to that finite family. There are pairwise disjoint chosen balls $B_1,\dots ,B_q$ among it such that $K\subseteq {\bigcup }_{j=1}^{q}5B_j\mathrm{.}$ Hence $\lambda (K)\le 5^n{\sum }_{j=1}^q\lambda (B_j)<5^{n}k{\sum }_{j=1}^q{\nu }_s(B_j)\le 5^{n}k{\nu }_s(U_{\epsilon })<5^{n}k\epsilon \mathrm{.}$

Since $\epsilon >0$ was arbitrary, step 3.1 gives $\lambda (K)=0$ for every compact $K\subseteq F_k$. Because $F_k$ is Borel by step 2.2, [L4] implies $\lambda (F_k)=\sup \{\lambda (K):K\subseteq F_k,K\text{ compact}\}=0.$ Step 2.2 now shows that the set where $L(x)>0$ is contained in the null set $({\mathbb{R}}^n\setminus G)\cup {\bigcup }_{k\ge 1}{F}_k\mathrm{.}$ The ratios defining $L$ are nonnegative, so ${\lim }_{r\to 0^{+}}\frac{{\nu }_s(B(x,r))}{\lambda (B(x,r))}=0$ for almost every $x$.

Combine step 4.1 with the comparison in step 2.1 and the differentiation theorem [L2] for the locally integrable representative $f$. For any specified Borel families $(E_r(x){)}_{r>0}$ shrinking nicely to the points of $A$, this gives $\frac{\nu (E_r(x))}{\lambda (E_r(x))}\to f(x)$ for almost every $x\in A$. Taking $A={\mathbb{R}}^n$ and $E_r(x)=B(x,r)$ gives the ball conclusion.