Given a positive, finite, regular measure $\lambda$ and $g\in L^1(\lambda)$, the measure $\mu$ given by $\mu(E)=\int_E g~d\lambda$ is regular

Question

Suppose $\lambda$ is a positive, finite, regular measure, and suppose $g \in L^1(\lambda)$. Define a measure $\mu$ by letting $\mu(E)=\int_E g~d\lambda$. Then is it true that $\mu$ is regular(i.e., $|\mu|$ is regular)?

This should be true, but I have no idea to show this. This question arises from the proof of Theorem 6.19 of Rudin's Real and Complex Analysis. He asserted that $\mu$ is regular without proof, but I cannot see why.

Answer 1

Let $\epsilon >0$. Since $\mu << \lambda$ there exists $\delta >0$ such that $\lambda (E) <\delta$ implies $|\int_E gd\lambda| <\epsilon$. Let $E$ be any Borel set and $K$ be a compact subset of $E$ with $\lambda (E\setminus K) <\delta$. Then $|\mu (E\setminus K)| \leq \int_{E\setminus K} |g| d\lambda <\epsilon$.