Stein 1.25

An alternative definition of measurability is as follows: $E$ is measurable if $\forall \epsilon > 0$, $ \exists$ closed set $F \subseteq E$ s.t. $m_*(E-F) \leq \epsilon$ $\star$. Show that this definition is equivalent (in the sense that each definition implies the other) to the definition given earlier in the text:

$E$ is meas. if $\forall \epsilon > 0$, $\exists \mathcal{O}_{open} \supseteq E$ s.t. $m_*(\mathcal{O}-E) \leq \epsilon$ $\star\star$

$\textit{Proof}$ : Let $\mathcal{M}$ be the set of Lebesgue measurable sets according to the definition $\star\star$ given in the text. Let $\mathcal{M}'$ be the collection of measurable sets given by $\star$. Shwoing that the two definitions are equivalent amounts to showing that $\mathcal{M} \subseteq \mathcal{M}'$ and $\mathcal{M}' \subseteq \mathcal{M}$.

"$\subseteq$" Let $E\in{\mathcal{M}}$. Then, since $\mathcal{M}$ is a $\sigma-$algebra, $E^c \in{\mathcal{M}}$. So, $\forall \epsilon >0$, $\exists \mathcal{O}_{open} \supseteq E^c$ s.t. $m_*(\mathcal{O}-E^c) \leq \epsilon$. $\mathcal{O}^c$ is closed since $\mathcal{O}$ is open. So, $\mathcal{O} \supseteq E^c \Longrightarrow \mathcal{O}^c \subseteq E$. If we can show that $E-\mathcal{O}^c \subseteq \mathcal{O}-E^c$ then we are done since the monotinicity of outer measure then implies that $m_*(E-\mathcal{O}^c) \leq m_*(\mathcal{O}-E^c) \leq \epsilon$.
Let $x\in{(E-\mathcal{O}^c)} \Longleftrightarrow x\in{E}$ and $x\not \in{\mathcal{O}^c} \Longleftrightarrow x\in{E}$ and $x\in{\mathcal{O}}$ $ \Longleftrightarrow x\in{\mathcal{O}}$ and $x\not \in{E}^c \Longleftrightarrow x\in{(\mathcal{O}-E^c)}$
Thus, $ \mathcal{M} \subseteq \mathcal{M}'$

"$\supseteq$" Let $E \in{\mathcal{M}'} \Longrightarrow \forall \epsilon >0 \quad \exists F_{closed} \subseteq E$ s.t. $m_*(E-F) \leq \epsilon$. Now, $F \subseteq E \Longrightarrow E^c \subseteq F_{open}^c$. If we can show that $F^c - E^c \subseteq E-F$ then the monotinicity of outer measure again implies $m_*(F^c-E^c) \leq m_*(E-F) \leq \epsilon$.
Let $x\in{(F^c-E^c)} \Longleftrightarrow x\not \in{F}$ and $x\not \in{E}^c \Longleftrightarrow x\in{E}$ and $x\not \in{F} \Longleftrightarrow x\in{(E-F)}$. By definition $\star$ this implies that $E^c \in{\mathcal{M}}$ which implies $E \in{\mathcal{M}}$.
Thus, $\mathcal{M} \supseteq \mathcal{M}'$.
$\therefore \mathcal{M}=\mathcal{M}'$ $\blacksquare$