Proposition 4.14(1) from Bass, http://bass.math.uconn.edu/3rd.pdf:
Suppose $A \subseteq [0, 1]$ is a Borel measurable set.
Let $m$ be Lebesgue measure.
(1) Given $\varepsilon>0$, there exists an open set $G$ so that $m(GāA) < \varepsilon$
and $A \subseteq G$.
My question is about the first sentence in the proof:
There exists a set of the form $E =\bigcup_{j=1}^\infty (a_j , b_j]$ such that $A ā E$ and $m(E ā A) < \varepsilon/2$.
What is the argument that such a union exists? The problem is to assure that $m(E ā A) < \varepsilon/2$.
I see some connection with the definition of Lebesgue outer measure
or perhaps I should prove the sentence
Every set in the Borel $\sigma$-algebra on $\mathbb{R}$ is $\mu^*$-measurable.
that I have found in my notes (without proof).
Best Answer
By definition of outer measure, you have $$m(A)= \inf \left\{ \sum_j (b_j-a_j) : A \subseteq \bigcup_j (a_j, b_j] \right\}$$ Fix $\varepsilon >0$. Then there exists a countable union of intervals $$E = \bigcup_j (a_j, b_j] \supseteq A$$ such that $$m(E)-m(A) < \varepsilon/2$$ Hence, since both $A$ and $E$ are Borel measurable, you have $$m(E-A) = m(E)-m(A) < \varepsilon/2$$ This proves the first claim.
To prove the second claim, i.e. every Borel is measurable, it is enough to prove that every open bounded interval $(a,b)$ is measurable. A proof of this can be found at Prove that any interval is measurable. , for example.
Then, since measurable sets form a sigma-algebra, countable unions of open intervals (read all open sets) are measurable. Again, since Borel sets are the sigma-algebra generated by open sets, it is contained in the sigma-algebra of all measurable sets.