Differentiation of sigma-finite Borel measures finite on compact sets

The closed balls $X_j=\overline{B(0,j)}$, $j\ge 1$, give a common finite-measure Borel exhaustion for $\nu$ and $\lambda$ by [L3]. Under the given AC, [L1] therefore gives $\nu ={\nu }_a+{\nu }_s$ and a finite real Borel density $f$ of ${\nu }_a$. Choose a Borel null set $N$ carrying ${\nu }_s$. For every Borel $E$, absolute continuity and concentration give ${\nu }_s(E)=\nu (E\cap N)\ge 0,{\nu }_a(E)=\nu (E\setminus N)\ge 0.$ Both components are positive and dominated by $\nu$. On each finite exhaustion piece, the set where $f\le -1\mathrm{/}k$ has measure zero: its density integral is at most $-\lambda (E)\mathrm{/}k$ but equals ${\nu }_a(E)\ge 0$. Taking the countable union over $j,k$ and replacing negative values by zero yields $f\ge 0$ without changing its class. For every ball, ${\int }_{B(x,R)}\mathrm{\mid }f\mathrm{\mid }d\lambda ={\nu }_a(B(x,R))\le \nu (\overline{B(x,R)})<\mathrm{\infty }\mathrm{.}$ Thus $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$.

We prove locally the outer approximation needed below. Let $\eta$ be any Borel measure finite on compact sets and $V=B(0,j)$ for $j\ge 1$; then $\eta (V)<\mathrm{\infty }$. Every open $W\subseteq V$ is an increasing union of compact sets: use the sets of points of $W$ with norm at most $m$ and distance at least $1\mathrm{/}m$ from its complement, $m\ge 1$. These are closed and bounded, so [L3] and [L5] approximate $W$ from inside by compact sets in $\eta$-measure. Let $\mathcal{C}$ be the Borel $E\subseteq V$ such that for every $\delta >0$ there are relatively closed $F$ and relatively open $O$ with $F\subseteq E\subseteq O\subseteq V$ and $\eta (O\setminus F)<\delta$. All relatively open sets are in $\mathcal{C}$ by the compact approximation just proved. Relative complements preserve $\mathcal{C}$ by interchanging $F$ and $O$. Countable unions preserve outer approximation by choosing open supersets with summable errors; they preserve inner approximation by first taking a finite partial union with small remaining measure (using $\eta (V)<\mathrm{\infty }$ and [L5]) and then taking a finite union of relatively closed approximants. Thus $\mathcal{C}$ is a sigma-algebra, and [L7] implies that it contains every Borel subset of $V$. In particular each such set has arbitrarily accurate relatively open outer approximations. The countable selections are covered by AC.

If $H$ is a Borel $\eta$-null set, put $V_0=\mathrm{\varnothing }$, $V_j=B(0,j)$ and $H_j=H\cap (V_j\setminus V_{j-1})$ for $j\ge 1$. Step 1.2 gives relatively open $O_j\subseteq V_j$ containing $H_j$ with $\eta (O_j)<\epsilon 2^{-j}$. As $V_j$ is open, each $O_j$ is open in ${\mathbb{R}}^n$. Their union is an open set containing $H$ with measure less than $\epsilon$. This proves the required null-set outer approximation without importing a later regularity theorem.

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 $\epsilon >0$. By steps 1.1 and 2.1, there is an open $U_{\epsilon }\supseteq G$ with ${\nu }_s(U_{\epsilon })<\epsilon$. Let $K\subseteq F_k$ be nonempty and compact (the empty case has measure zero). Consider all balls $B(x,r)\subseteq U_{\epsilon }$ with $x\in K$ and ${\nu }_s(B(x,r))>\lambda (B(x,r))\mathrm{/}k$. They cover $K$: for each $x$, choose $m$ with $B(x,1\mathrm{/}m)\subseteq U_{\epsilon }$, and use $D_m(x)>1\mathrm{/}k$. Compactness gives a finite subcover. Greedily retain a largest-radius ball and discard all balls meeting it, repeating on the remaining finite family. The retained balls $B_1,\dots ,B_q$ are disjoint. Any discarded ball of radius $r$ meets a retained one of radius $R\ge r$; their centres are less than $r+R$ apart, so every point of the discarded ball is less than $2r+R\le 3R<5R$ from the retained centre. Hence $K\subseteq {\bigcup }_{j}5B_j$. By [L6] and disjointness, $\lambda (K)\le 5^n{\sum }_j\lambda (B_j)<5^{n}k{\sum }_j{\nu }_s(B_j)\le 5^{n}k{\nu }_s(U_{\epsilon })<5^{n}k\epsilon \mathrm{.}$

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 2.3, [L4] implies $\lambda (F_k)=\sup \{\lambda (K):K\subseteq F_k,K\text{ compact}\}=0.$ Step 2.3 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 2.2 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.