Characterization of measurability by closed sets.

Question

If $E \subseteq \Bbb R$ is measurable, then for all $\epsilon > 0$, exists $F \subseteq \Bbb R$ closed such that $F \subseteq E$ and ${\frak m}^*(E \setminus F) < \epsilon$.

I have already made the characterization by open sets, that is, if $E$ is measurable, for all $\epsilon > 0$, exists $O \subseteq \Bbb R$ open, such that $E \subseteq O$ and ${\frak m}^*(O \setminus E) < \epsilon$.

For that characterization, I used that for all $A \subset \Bbb R$, given $\epsilon > 0$, exists $O \subseteq \Bbb R$ open such that ${\frak m}^*O < {\frak m}^*A + \epsilon$. Just take intervals $(I_j)_{j \geq 1}$ covering $A$ such that $\sum_{j \geq 1} \ell (I_j) < {\frak m}^*A + \epsilon$, so let $O = \bigcup_{j \geq 1} I_j$ if ${\frak m}^*A < +\infty$, and $O = \Bbb R$ if ${\frak m}^*A = +\infty $.

I was trying to "copy" this proof, and so I would begin proving the lemma (which I firmly believe that is true):

For all $A \subseteq \Bbb R$, given $\epsilon > 0$, exists $F \subseteq \Bbb R$ closed such that ${\frak m}^*A < {\frak m}^*F + \epsilon$.

Well, if ${\frak m}^*A = 0$, take $F = \varnothing$. But apart from this, my attempt isn't going well. Surely, if ${\frak m}^*A > 0$, we can take intervals $(I_j)_{j \geq 1}$ covering $A$ such that $\sum_{j \geq 1} \ell (I_j) < {\frak m}^*A - \epsilon$, but so what? Infinite union of closed sets need not be closed.

Another idea is to use the result of open sets for $A^c$, so that $A^c \subseteq O \implies O^c \subseteq A $, then take $F = O^c$. But I don't see quickly how to get the relation between the measures.

Am I going in the right way? Can someone help me please?

Answer 1

We must show that:

If $E \subseteq \Bbb R$ is measurable, then for all $\epsilon > 0$, exists $F \subseteq \Bbb R$ closed such that $F \subseteq E$ and ${\frak m}^*(E \setminus F) < \epsilon$.

So let $E, \epsilon$ be as above. $E^c$ is measurable, and, as you know, there exists an open set $O$ such that $E^c \subset O$ and $m(O\setminus E^c)<\epsilon$.

Now let $F=O^c$. Then it's easy to see that $F \subset E$. All that remains to show is that $m(O \setminus E^c)=m(E\setminus F)$

Notice that $E \setminus F=E \setminus O^c$ is the same set as $O \setminus E^c$. If $x$ is an element of $E\setminus O^c$,then $x \in E$ and $x \notin O^c$. This means $x \notin E^c$ but $x \in O$, therefore $x \in O\setminus E^c$. Now suppose $x \in O\setminus E^c$. this means $x \in O$ and $x \notin E^c$, but this means $x \notin O^c$ and $x \in E$, therefore $x \in E\setminus O^c$.