Showing two measures are equal up to scalar multiplication.

Question

I've been stuck on the following question and was wondering if it'd be possible to get some help.

Let $\mu, \nu$ be two finite Borel measures on a metric space $X$ s.t. $\mu << \nu$ and let $c>0$ and suppose that $\nu$ is doubling and that $$\lim_{r \to 0} \frac{\mu(B_r(x))}{\nu(B_r(x))} = c$$ for $\nu$ almost every $x\in X$. By using the Vitali Covering theorem to show that $\mu = c\nu$. (We assume the balls are closed)

I've been able to prove this using Radon-Nikodym combined with the lebesgue differentiation theorem but wish to prove it solely using Vital's covering theorem but have been unsuccessful in even finding an appropriate Vitali cover.

Theorem(Vitali Covering Theorem)
Let X be a metric space and let $\nu$ be a doubling measure on $X$ and let $\mathcal{B}$ be a Vitali cover for $S \subset X$ then there exists a countable $\mathcal{B}' \subset \mathcal{B}$ such that all elements of $\mathcal{B}'$ are disjoint and $$\nu \bigg(S \setminus \bigcup_{B' \in \mathcal{B}'}B'\bigg)=0$$

Definition(Vitali Cover)
Let $S \subset X$ then $\mathcal{B}$ a collection of closed balls such that $\forall \epsilon >0 , \forall x \in S$ there exists $B \in \mathcal{B}$ such that $x \in B$ and $\rm{rad}(B) < \epsilon$ then we call $\mathcal{B}$ a Vitali cover for S.

Many Thanks.

Answer 1

Let's prove that for any $\mu$ and $\nu$ measurable $S$ in $X$ we have $\mu(S) = c\nu(S).$ Let $0 < \epsilon < 1.$

For $\mu$-a.e. $x \in S$ there exists $r_1(x) > 0 $ such that for all $r < r_1(x).$

\begin{equation} \mu(S \cap B(x,r)) \geq (1- \epsilon)\mu(B(x,r)). \tag{1} \end{equation}

For $\nu$-a.e. $x \in S$ there exists $r_2(x) > 0$ such that for all $r < r_2(x)$ we have

\begin{equation} \nu(S \cap B(x,r)) \geq (1- \epsilon)\nu(B(x,r)). \tag{2} \end{equation}

Note that (1) and (2) come from Lebesgue's density theorem which we can apply since $\mu$ and $\nu$ are finite doubling measures. Finally for $\nu$-a.e. $x \in X$ there exists $r_3(x) > 0$ such that for all $r < r_3(x)$ we have

$$\frac{c}{1 + \epsilon}\nu(B(x,r)) \leq \mu(B(x,r)) \leq c(1 + \epsilon) \nu(B(x,r)) \tag{3}.$$

Let $R(x) = \min(r_1(x), r_2(x), r_3(x))$ and let

$$\mathcal{B} = \{B(x,r) \mid r < R(x), B(x,r) \ \text{satisfies (1) (2) and (3)}\}.$$

This is an (a.e.) Vitali cover of $S,$ so we can apply the Vitali covering theorem to get a disjoint subset $\{B_1, B_2,...\}$ such that

$$\nu\bigg(S\setminus \bigcup_{i=1}^\infty B_i\bigg) = 0 = \mu\bigg(S\setminus \bigcup_{i=1}^\infty B_i\bigg),$$

with the second inequality following from $\mu \ll \nu.$ Since $S$ is measurable we deduce that

$$\mu(S) = \mu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg)$$

and

$$\nu(S) = \nu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg).$$

Applying all of the above we have

\begin{align} \mu(S) &= \mu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg) \\ &= \sum_{i=1}^\infty \mu(S \cap B_i) \\ &\leq \sum_{i=1}^\infty \mu(B_i) \\ &\leq c(1+\epsilon)\sum_{i=1}^\infty \nu(B_i) \\ &\leq \frac{c(1 + \epsilon)}{(1-\epsilon)}\sum_{i=1}^\infty \nu(S \cap B_i) \\ &=\frac{c(1 + \epsilon)}{(1-\epsilon)}\nu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg)\\ &= \frac{c(1 + \epsilon)}{(1-\epsilon)} \nu(S). \end{align}

Since $\epsilon$ was arbitrary it follows that $\mu(S) \leq c\nu(S).$ The reverse inequality is virtually identical giving that $\mu(S) = c\nu(S).$

Answer 2

It helps to know that finite Borel measures on a metric space $X$ are regular. In particular, if $E \subset X$ is a Borel set, $\mu$ is a finite Borel measure, and $\epsilon > 0$, then there exists an open set $G \supset E$ with $\mu(G) < \mu(E) + \epsilon$.

Assume that the stated conditions on $\mu$ and $\nu$ hold (and in particular $\nu(B(x,r)) > 0$ for all $x \in X$ and $r > 0$).

Let $E \subset X$ be a Borel set. Fix a constant $t > 1$.

For every $x \in E$ there exists $r_x > 0$ with the property that $0 < r < r_x$ implies $$\frac ct < \frac{\mu(B(x,r))}{\nu(B(x,r))} < ct,$$ which can be expressed as $$0 < r < r_x \implies \frac ct \nu(B(x,r)) \le \mu(B(x,r)) \le ct \nu(B(x,r)).$$

Let $\epsilon > 0$ and let $G \supset E$ be an open set with $\nu(G) < \nu(E) + \epsilon$. Define a Vitali cover ${\cal V}$ of $E$ by $$ {\cal V} = \{B(x,r) : x \in E,\ 0 < r < r_x,\ B(x,r) \subset G\}.$$ According to Vitali's covering theorem there exists a countable disjointed subfamily ${\cal V}'$ of $\cal V$ satisfying $$\mu(E) \le \mu \left( E \setminus \bigcup_{B \in {\cal V}'} B \right) + \mu \left(\bigcup_{B \in {\cal V}'} B \right) \le \sum_{B \in {\cal V}'} \mu(B) \le ct\sum_{B \in {\cal V}'} \nu(B)$$ Since the family ${\cal V}'$ is pairwise disjoint and each $B \in {\cal V}'$ is a subset of $G$ you have in addition that $$\sum_{B \in {\cal V}'} \nu(B) \le \nu(G) <\nu(E) + \epsilon.$$ Thus $\mu(E) < ct(\nu(E) + \epsilon)$ and by letting $\epsilon \to 0^+$ you get $$\mu(E) \le ct \nu(E).$$

Now let $\epsilon > 0$ and let $H \supset E$ be an open set with $\mu(H) < \mu(E) + \epsilon$. Repeat the steps above to find that $$\frac ct \nu(E) \le \mu(E)$$ and conclude $$\frac ct \nu (E) \le \mu(E) \le ct \nu(E).$$ Finally let $t \to 1^+$.