Real Analysis Folland, 1.20 Proposition Borel measures

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

Proof - Let $E\in M_{\mu}$ and $\mu(E) < \infty$ then by lemma 1.17 there exists an open set $U\in M_{\mu}$ such that $E\subseteq U$ and $\mu(U)\leq \mu(E) + \epsilon/2$. Let $\{U_n\}_{1}^{\infty}$ be a sequence of disjoint open intervals such that $\bigcup_{1}^{\infty}U_n = U$ Then $$\sum_{1}^{\infty}\mu(U_n) = \mu(U) < \infty$$ so there exists an $N\in\mathbb{N}$ such that $\sum_{n=N+1}^{\infty}\mu(U_n) < \epsilon/2$. Define $A:= \bigcup_{1}^{n} U_n$ It follows that $$\mu(E \ \triangle \ A) = \mu(E\setminus A) + \mu(A\setminus E)\leq \mu(E\setminus U) + \mu(U\setminus E) + \mu(U\setminus A) = 0 + \mu(U) - \mu(E) + \mu(U) - \mu(A) < \epsilon/2 + \epsilon/2 = \epsilon$$

I am not sure if this is correct, any suggestions or comments are greatly appreciated.

• What is your question? Jun 13 '16 at 22:29
• @WillKwon edited my question Jun 13 '16 at 22:30
• It's correct, but why not just $\mu(E\setminus A) + \mu(A\setminus E) \leq \mu(U\setminus A) + \mu(U\setminus E)$. Jun 13 '16 at 22:42
• Idea is almost correct. But there are some minor mistakes. First, the first identity, we cannot gurantee that $E\setminus A$ and $A\setminus E$ are disjoint. Also, the first inequality does not hold. If you follow the Qiyu Wen's comment, then you will get a precise proof. Jun 13 '16 at 22:49

1.20 Proposition - If $E\in M_{\mu}$ and $\mu(E) < \infty$, then for every $\epsilon > 0$ there is a set $A$ that is finite union of open intervals such that $\mu(E \ \triangle \ A) < \epsilon$.
Proof - Let $E\in M_{\mu}$ and $\mu(E) < \infty$ then by lemma 1.17 there exists an open set $U\in M_{\mu}$ such that $E\subseteq U$ and $\mu(U)\leq \mu(E) + \epsilon/2$. Let $\{U_n\}_{1}^{\infty}$ be a sequence of disjoint open intervals such that $\bigcup_{1}^{\infty}U_n = U$ Then $$\sum_{1}^{\infty}\mu(U_n) = \mu(U) < \infty$$ so there exists an $N\in\mathbb{N}$ such that $\sum_{n=N+1}^{\infty}\mu(U_n) < \epsilon/2$. Define $A:= \bigcup_{1}^{N} U_n$ It follows that
Remark: The fact that $\mu(E) <\infty$ is used to ensure that $\mu(U\setminus E) =\mu(U) - \mu(E)$.