# Why $\subseteq$ and not $=$ in the proof of theorem 11 on pg. 41 in Royden “real analysis ” fourth edition

analysiselementary-set-theorylebesgue-measuremeasure-theoryreal-analysis

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?

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.