I am going over Folland’s excellent book Real Analysis: Modern Techniques and Their Applications so I can brush on my measure theory skills.

I have been a bit lazy to go over all exercises. Therefore, I decided to commit at least one exercise every day on this blog hoping it will serve as a good motivator. I want to stress that these are my own solutions, so if you are counting on using them, do them please at your own risk.

So, without further ado, I present to you problem 1.25:

Prove Proposition 1.20. If $E \in \mathcal{M}_\mu$ and $\mu(E)<\infty$, then for every $\epsilon>0$ there is a set $A$ that is a finite union of open intervals such that $\mu(E \Delta A)<\epsilon$.

**Solution**. Pick an $\epsilon>0$. We know from **Theorem 1.18** that there is an open set $U$ such that

$$ \mu(U) \le \mu(E) + \epsilon,$$ and $E \subset U$. We also know that $$ \mu(U-E)=\mu(U\triangle E)=\mu(U)-\mu(E) \le \epsilon,$$ where I’ve used the fact that $\mu(E)<\infty$ and that $E \subset U$. This implies that $\mu(U)< \infty$.

Since $U$ is an open set, we know that in $\mathbb{R}$, $U$ can be represented as the union of at most countable disjoint open sets. Since $\mu(U)<\infty$, we have that it is the union of finite disjoint open sets $\blacksquare$