I have a question about a proof of the following statement (the Lebesgue measure is defined with half-open intervals).
Suppose $A\subset [0,1]$ is a Lebesgue measurable set. Let $m$ be the Lebesgue measure. Given $\varepsilon>0$, there exists an open set $G$ so that $m(G-A)<\varepsilon$ and $A\subset G$.
There exists $E=\bigcup_{j=1}^{\infty}(a_j,b_j]$ such that $A\subset E$ and $$m(E)\leq \sum_{i=1}^\infty (b_j-a_j)<m(A)+\frac12\varepsilon$$
via the definition of the Lebesgue measure and countable subadditivity. We then note that since $A\subset E$, $m(A-E)=m(A)-m(E)$, so $m(A-E)<\frac12\varepsilon$.
Now, let $G=\bigcup_{j=1}^\infty (a_j,b_j+\varepsilon 2^{-j-1})$. In fact, $G$ is open and contains $E$ and $A$. I want to show that
$$m(G-E)\leq\sum_{j=1}^\infty \varepsilon2^{-j-1}=\frac12\varepsilon$$
and conclude since $m(G-A)=m(G-E)+m(E-A)$.
However, I am having trouble showing this step. I have that
$$m(G)\leq \sum_{j=1}^\infty (b_j-a_j)+\sum_{j=1}^\infty \varepsilon2^{-j-1}$$
but I am not sure how to combine it with another inequality like
$$m(E)\leq \sum_{j=1}^\infty (b_j-a_j)$$
to get the result that I want. I think I'm pretty close, but I am not sure how to proceed.
Best Answer
Use the fact that $G\setminus E \subseteq \cup_{j=1}^{\infty} (b_j, b_j+\epsilon /2^{j+1})$ and hence $m(G\setminus E ) \leq \sum\limits_{j=1}^{\infty} \frac {\epsilon} {2^{j+1}}=\frac {\epsilon} {2}$.