# Open set approximating measurable set

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 $$m(E)\leq \sum_{i=1}^\infty (b_j-a_j) via the definition of the Lebesgue measure and countable subadditivity. We then note that since $$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.

Use the fact that $$G\setminus E \subseteq \cup_{j=1}^{\infty} (b_j, b_j+\epsilon /2^{j+1})$$ and hence $$m(G\setminus E ) \leq \sum\limits_{j=1}^{\infty} \frac {\epsilon} {2^{j+1}}=\frac {\epsilon} {2}$$.