Lemma 2.8.19790D3B21542

Proof. We first construct the density on a half-open cube as an $L^2$ limit of averages of $\mu $ over dyadic subcubes. Let $Q=\prod _{i=1}^ d[a_ i,a_ i+s)$ be a half-open cube whose closure lies in $\Omega $, so that $\nu (Q)<\infty $, and suppose that $\mu (E)\leq C\nu (E)$ for Borel sets $E\subset Q$; in particular, $\mu (Q)\leq C\nu (Q)<\infty $. For $k\geq 0$, let $\mathcal Q_ k$ be the partition of $Q$ into $2^{kd}$ half-open dyadic subcubes of side $2^{-k}s$. On each $Q'\in \mathcal Q_ k$, let $g_ k=\mu (Q')/\nu (Q')$ if $\nu (Q')>0$ and $g_ k=0$ otherwise; in the latter case also $\mu (Q')=0$, by domination. Thus $0\leq g_ k\leq C$, and $\int _{Q'}g_ l\, \mathrm d\nu =\mu (Q')$ whenever $Q'\in \mathcal Q_ k$ and $l\geq k$, because $Q'$ is a disjoint union of cubes of $\mathcal Q_ l$. As $g_ k$ is constant on the cubes of $\mathcal Q_ k$, it follows that $\int _ Qg_ lg_ k\, \mathrm d\nu =\int _ Qg_ k^2\, \mathrm d\nu $, and hence, in $L^2(Q,\nu )$,

The squared norms increase and are bounded by $C^2\nu (Q)$, so $(g_ k)$ is a Cauchy sequence in $L^2(Q,\nu )$. Choose $k_1<k_2<\cdots $ with $\| g_{k_{j+1}}-g_{k_ j}\| _2\leq 2^{-j}$. By the Cauchy–Schwarz inequality, $\int _ Q|g_{k_{j+1}}-g_{k_ j}|\, \mathrm d\nu \leq \nu (Q)^{1/2}2^{-j}$, so by monotone convergence $\sum _ j|g_{k_{j+1}}-g_{k_ j}|$ has finite integral; hence it is finite, and $(g_{k_ j})$ converges, $\nu $-almost everywhere. Let $g$ be the limit of $(g_{k_ j})$ on the Borel set where it converges, and $0$ elsewhere. Then $g$ is a Borel function with values in $[0,C]$, $g_{k_ j}\to g$ in $L^2(Q,\nu )$ by dominated convergence, and, the sequence being Cauchy, $g_ k\to g$ in $L^2(Q,\nu )$ and hence in $L^1(Q,\nu )$. Consequently $\int _{Q'}g\, \mathrm d\nu =\mu (Q')$ for every dyadic subcube $Q'$ of $Q$.

The dyadic subcubes of $Q$, together with $\varnothing $, form a family closed under finite intersections, since two of them are nested or disjoint. If $U$ is open, every point of $U\cap Q$ lies in arbitrarily small dyadic subcubes, hence in one contained in $U$, so $U\cap Q$ is a countable union of dyadic subcubes. Thus these cubes generate the Borel $\sigma $-algebra of $Q$, and lemma 2.2.6(2) shows that the finite measures $\mu $ and $g\nu $, which agree on all dyadic subcubes including $Q$, agree on every Borel subset of $Q$.

For uniqueness, let $g,h\geq 0$ be Borel functions with $g\nu =h\nu =\mu $ on the Borel subsets of such a cube $Q$. The set $A=Q\cap \{ g>h\} $ satisfies $\int _ A(g-h)\, \mathrm d\nu =\mu (A)-\mu (A)=0$, because $\mu (A)<\infty $, so $\nu (A)=0$; by symmetry, $g=h$ $\nu $-almost everywhere on $Q$. Since $\Omega $ is a countable union of such cubes, densities on $\Omega $ are unique up to $\nu $-null sets.

To construct the density on $\Omega $ in case (1), cover $\Omega $ by countably many half-open cubes $Q_1,Q_2,\ldots $ with closures in $\Omega $, for instance all half-open dyadic cubes with this property, let $g^{(k)}$ be the density on $Q_ k$ just constructed, and put $g=g^{(k)}$ on $P_ k=Q_ k\setminus \bigcup _{l<k}Q_ l$. For every Borel set $E\subset \Omega $,

so $\mu =g\nu $ with $0\leq g\leq C$. In case (2), every cube $Q$ with closure in $\Omega $ lies in a relatively compact open set $U\subset \Omega $, for instance a set of points at distance less than $\delta $ from $\overline Q$ with $\delta $ small, and $\mu \leq C_ U\nu $ on Borel subsets of $Q$; the same construction gives $\mu =g\nu $ with $g\geq 0$ Borel. If $U\subset \Omega $ is relatively compact and open, the sets $A_ j=U\cap \{ g>C_ U+1/j\} $ satisfy $(C_ U+1/j)\nu (A_ j)\leq \mu (A_ j)\leq C_ U\nu (A_ j)$ and $\nu (A_ j)<\infty $, so $\nu (A_ j)=0$; thus $g\leq C_ U$ $\nu $-almost everywhere on $U$. The final assertion is case (1) for Lebesgue measure, which is a Radon measure by proposition 2.2.2. $\square$