I am trying to solve the following qual problem:
Given a finite Borel measure $\mu$ on $\mathbb{R}$, its support is the set $$S:= \{x \in \mathbb{R}: \mu((x-\epsilon,x + \epsilon)>0 \, \text{for every } \epsilon >0\}.$$ Prove that $S$ is closed and $\mu(\mathbb{R}/S) = 0$. Moreover, any other set with these two properties must contain $S$.
Prove that there is a finite Borel measure $\mu$ on $\mathbb{R}$ such that
- $\mu$ has support equal to $\mathbb{R}$
- $\mu$ and Lebesgue measure are mutually singular.
(UCLA, Spring 2022 Qual)
My attempt:
Firstly, we want to show that $S$ is closed. Let $\{x_n\}$ be a convergent sequence in $S$ whose limit is $x$. Our aim is to show that $x \in S$. Given $\epsilon >0$, we need to show that $\mu((x-\epsilon,x + \epsilon))>0$. Since, $\{x_n\}$ converges to $x$, we can find $M \in \mathbb{N}$ such that $|x_n-x| < \epsilon /2$ for all $n > M$. As a result, $(x_n - \epsilon/2 , x_n + \epsilon / 2) \subset (x - \epsilon ,x + \epsilon)$. Thus, $\mu((x-\epsilon,x + \epsilon))>0$ follows. Hence, $S$ is closed.
Secondly, we need to show that $\mu(\mathbb{R} /S) = \mu(S^c)=0$. For each $x \in S^c$, by definition, there exist $\epsilon_x >0$ such that $\mu(x - \epsilon_x , x + \epsilon_x)=0.$ Intuitively, I try to show that $S^c$ can be written as a countable union of intervals of measure $0$.(I am not sure whether this is true or not.)
Also, I do not know how to show that $S$ is the "smallest" set satisfying these properties.
Thirdly, for constructing a Borel measure with the given properties, I suspect we need to use some sort of measure decomposition theorems. However, I do not know how to do it properly.
Any help would be appreciated, thanks in advance.
If $x \in \mathbb R \setminus S$ then there exist rational numbers $r$ and $s$ such that $x \in (r,s)$ and $\mu ((r,s))=0$. $\mathbb R \setminus S$ is covered by these intervals and a countable union of sets of measure $0$ has measure $0$. Hence, $\mu (\mathbb R \setminus S)=0$.
Let $(r_n)$ be an ennumeration of the set of all rational numbers and $\mu (E)=\sum\limits_{n=1}^{\infty} 2^{-n}\chi_E(r_n)$. Then $\mu$ is a probability measure and $\mu (a,b) \gt 0$ whenever $a \lt b$. Can you show from this that the support of $\mu$ is $\mathbb R$? Also, $\mu(\mathbb Q)=1$ so $\mu$ is singular w.r.t. Lebegue measure.
Suppose $T$ is a closed set such that $\mu (\mathbb R \setminus T)=0$. If $x \notin T$ then (since $\mathbb R \setminus T$ is open) there exist $\epsilon >0$ such that $\mu (x-\epsilon, x+\epsilon)=0$. Hence, $x \notin S$. We have proved that $S \subseteq T$,