Here is the theorem:

Here is a part of the proof:

First, We will establish the equivalence of the measurability of $E$ with each of the two outer approximation properties $(i)$ and $(ii).$

Assume that $E$ is measurable. Let $\epsilon > 0.$ First consider the case $m^*(E) < \infty.$ By the definition of $m^*,$ there is a countable collection of nonempty, bounded, open intervals $\{I_k\}_{k=1}^{\infty}$ which covers $E$ and for which $$\sum_{k=1}^{\infty} \ell(I_k) < m^*(E) + \epsilon $$

Define $\mathcal{O} := \cup_{k=1}^{\infty} I_k.$ Then $\mathcal{O}$

is an open set containing $E.$ By the definition of the outer measure of $\mathcal{O},$ we have $$m^*(\mathcal{O}) \leq \sum_{k=1}^{\infty} \ell(I_k) < m^*(E) + \epsilon $$

So that $$m^*(\mathcal{O}) – m^*(E) < \epsilon $$

However, $E$ is measurable and has finite outer measure. Therefore, by the excision property of measurable sets, we have $$m^* (\mathcal{O} \setminus E) = m^*(\mathcal{O}) – m^*(E) < \epsilon$$ as required.\

Now, consider the case when $m^*(E) = \infty.$ Then $E$ may be expressed as the disjoint union of a countable collection $\{E_k\}_{k=1}^{\infty}$ of measurable sets, each of which has finite outer measure. Now, by the finite measure case (proved above), for each $k,$ there is an open set $\mathcal{O}_k$ containing $E_k$ for which $$m^* (\mathcal{O}_k \setminus E_k) = m^*(\mathcal{O}_k) – m^*(E_k) < \epsilon / 2^k.$$ The set $\mathcal{O} = \cup_{k=1}^{\infty} \mathcal{O}_k$ is open, it contains $E$ and $$\mathcal{O} \setminus E = \bigcup_{k=1}^{\infty} \mathcal{O}_k \setminus E \subseteq \bigcup_{k=1}^{\infty} [\mathcal{O}_k \setminus E_k]$$

**What confuses me is the last part in the last line, namely**

$$\mathcal{O} \setminus E = \bigcup_{k=1}^{\infty} \mathcal{O}_k \setminus E \subseteq \bigcup_{k=1}^{\infty} [\mathcal{O}_k \setminus E_k]$$

Why the last step is $\subseteq$ and not $=$, could anyone explain this to me please?

## Best Answer

If you mean $\mathcal{O}_k\setminus E_k$ in the last expression (which i believe is the correct expression in order to use the fact that $m^*(\mathcal{O}_k\setminus E_k)<\varepsilon/2^k$), then the idea is that in $\bigcup\mathcal{O}_k\setminus E$ you are removing the whole set $E$ out of each $\mathcal{O}_k$, while in $\bigcup [\mathcal{O}_k\setminus E_k]$ you only remove a part of $E$ from each $\mathcal{O}_k$, namely $E_k$. So equality only holds if $\mathcal{O}_k\setminus E_k=\mathcal{O}_k\setminus E$ for each $k$, and this is not always the case.

For the question as it stands, note that $E$ has infinite outer measure, but $E_1$ has finite outer measure, so $E_1\subsetneq E$. Let $x\in E\setminus E_1$. As $E\subset\mathcal{O}$, we have $$x\in \mathcal{O}\setminus E_1\subset \bigcup_k [\mathcal{O}\setminus E_k],$$ but $x\notin \mathcal{O}\setminus E$, so equality cannot hold.