(T/F) If $\mu$ is a Borel measure on $\mathbb{R}$ and $A$ is a Borel set such that $\mu(A \cap K) = 0$ for all compact sets $K$, then $\mu(A) = 0$.

Question

I am trying to prove or disprove the following statement:

If $\mu$ is a Borel measure on $\mathbb{R}$ and $A$ is a Borel set such that $\mu(A \cap K) = 0$ for all compact sets $K$, then $\mu(A) = 0$.

I am looking for intuition behind the statement, since I am still trying to understand the Borel sigma-algebra.

Answer 1

$$ \mu(A)=\lim_{n\to\infty}\mu(A\cap[-n,n])=0. $$

Answer 2

Note by countable additivity of measure,

$0 \leq \mu(A)= \sum_{n\in \mathbb Z} \mu(A \cap (n,n+1]) \leq \sum_{n\in \mathbb Z} \mu(A \cap [n,n+1])= 0$

Hence $\mu(A)=0$.