# Inner Approximation by Closed Sets.

There is a theorem in our textbook that says:

Suppose that $E$ is measurable. "For each $\epsilon > 0$ , there is a closed set $F$ contained in $E$ for which $m*(E~F) \lt \epsilon$."

My question:

Since E is measurable, we know that,

$m^*(F) = m^*(F \cap E) + m^*(F \cap E^c) \rightarrow m^*(F \cap E^c) = m^*(F) - m^*(F \cap E)$ But since F is contained in E, $m^*(F) = m^*(F \cap E)$ So isn't $m^*(E \sim F)=m^*(F \cap E^c) = 0$? Then why do we need to consider epsilon?

$E-F$ is not the same as $F\cap E^c=F-E$.
Then what is $E - F$? – Idonknow Apr 29 '15 at 7:39