Let $\mu$ be a measure on $(\mathbb{R},\mathcal{B})$. Let $B \in \mathcal{B}$ be such that $B \in (-\infty,\infty) $ and $\mu(B) >0$. Give a proof or else give a counter example to the assertion: There exists a point $x \in B$ such that $$ \mu((x-\delta,x+\delta)) > 0 \:\:\forall \delta >0 $$
My thoughts
I think the statement is true, and therefore I am trying to prove it. Clearly, $B$ is an uncountable union of all such intervals. I am trying to find a countable sub cover. I'll appreciate a help.
I'm assuming $\mathcal{B}$ are the Borel sets (otherwise it doesn't need to be true). Suppose $B\in\mathcal{B}$ is a set such that for all then for all $x\in B\,,$ there exists a $\delta_x>0$ such that $\mu((x-\delta_x,x+\delta_x)=0\,.$ Now these sets cover $B$ so since $\mathbb{R}$ is Lindelöf we can extract a countable subbcover $\{(x_k-\delta_k,x_k+\delta_k)\}_k\,,$ hence we have $0\le\mu(B)\le \sum\limits_{k=0}^\infty\mu(x_k-\delta_k,x_k+\delta_k)=0\,,$ so that $\mu(B)=0\,.$ This proves the contrapositive, proving your claim.