I'm studying Measure theory at university. For practising with some exercises, I found some available notes on the web. While reading chapter about Borel measures, the following definitions are given:
A Borel measure $\mu$ on the topological space $X$ is a measure defined on a $\sigma$-algebra that contains the Borel $\sigma$-algebra;
A Borel measure $\mu$ on the topological space $X$ is called regular if, for each Borel set $E$ in $X$ $\mu(E)=\inf\{\mu(U) | U \text{ open}\supseteq E \}$,
$\mu(E)=\sup\{\mu(K)|K\text{ compact}\subseteq E\}$
Given this context, at the end of the chapter, I found the following exercise I cannot solve:
Let $\mu$ be a $\sigma$-finite regular Borel measure on the topological space $X$ and $Y$ be a Borel subset of $X$. Prove that the restriction $\mu_Y$ is a regular Borel measure on $X$. Here $\mu_Y$ is a measure defined on the same $\sigma$-algebra of $X$ such that $\mu_Y(E):=\mu(E\cap Y)$ for each set $E$ in the $\sigma$-algebra.
I think a good strategy can be to solve the exercise for Borel sets $E$ such that $\mu(E)<\infty$, and then to exploit the $\sigma$-finiteness of $\mu$ to achieve the general result.
Any idea on how to proceed?