# Question on Lebesgue measure and outer measure

Let $$m^*$$ denote the Lebesgue outer measure and when the set is measurable, define Lebesgue measure $$m:=m^*$$.

Prove the following:

Take any set $$E\subset\mathbb R^n$$ (not necessarily measurable), for any $$\varepsilon>0$$, there exist open (thus measurable) set $$U$$, such that $$m(U)\leqslant m^*(E)+\varepsilon$$

I think if we can construct a finite open cover of $$\partial E$$, i.e. $$\partial E\subseteq \bigcup_{j\in J} W_j$$, such that $$m(W_j)<\frac{\varepsilon}{|J|}$$ (?) Then define $$U:=\text{int}(E)\cup\left( \bigcup_{j\in J}W_j\right)$$, it is open because it is a finite union of open sets. And $$U=E\cup\left( \bigcup_{j\in J}W_j\right)$$, so by finite subaddativity: $$m(U)\leqslant m^*(E)+\sum_{j\in J}m(W_j)

Is the construction (?) really possible? If so, how to make a rigorous argument? The reason why I believe it works is that $$m^*(E)=m^*(\overline{E})$$

Thanks!

• Are you sure about $m^*(E)=m^*(\overline E)$? What if $E$ is a countable dense subset of $\mathbb R^n$? – bof May 11 '19 at 8:16

We are not concerned with the lebesgue measureability of E because we won’t be taking an actual measure on it, just $$m^*$$. Don’t worry about the boundary of E whatsoever, and chase the definition of $$m^{*}$$, which is the infemum of the sum of measures of collections of balls (or open sets) which cover E. By definition, if $$\epsilon$$ > 0, there exists a countable collection of open sets $${O_i}$$ with $$E \subset \cup_i O_i$$ and $$m^*(E) \leq \Sigma_i m(O_i) < m^*(E)+ \epsilon$$
Then $$O = \cup_i O_i$$ is an open measureable set, with $$E \subset O$$ so that, by subadditivity $$m^*(E) \leq m^*(O) = m(O) \leq \Sigma_i m(O_i) < m^*(E)+\epsilon$$.