Show that for each $A\subseteq \mathbb{R}$ there is a Borel subset $B\subseteq \mathbb{R}$
that includes $A$ and Lebesgue measure of $B$ is equal to outer Lebesgue measure of $A$.
I'd like to prove it by using only definition of Lebesgue outer measure for $A\subseteq \mathbb{R}$:
$\lambda^*(A)=\inf\{ \sum_i(b_i-a_i): A \subseteq \bigcup<a_i,b_i> \}$
I would be grateful for detailed explanation!
I tried proving that I could replace union with open sets (that's not problem to prove) but I'm stuck with how to prove that I can replace this infinite sum with measure of that open sets.
Best Answer
Suppose $\lambda^{\ast}(A) = \infty$, take $B = \mathbb{R}$.
If $\lambda^{\ast}(A) < \infty$, then for any $k \in \mathbb{N}$, $\lambda^{\ast}(A) + 1/k$ is not a lower bound for the set $$ \{\sum (b_i - a_i) : A \subset \cup (a_i, b_i)\} $$ Hence, there is a set $B_k := \cup (a_i, b_i)$ such that $A\subset B_k$ and $$ \lambda^{\ast}(B_k) = \sum (b_i - a_i) < \lambda^{\ast}(A) + 1/k $$ Take $B = \cap_k B_k$, then $A\subset B$, so $\lambda^{\ast}(A) \leq \lambda^{\ast}(B)$, and also $$ \lambda^{\ast}(B) \leq \lambda^{\ast}(B_k) \leq \lambda^{\ast}(A) + 1/k \quad\forall k\in \mathbb{N} $$ Hence, $\lambda^{\ast}(A) = \lambda^{\ast}(B)$ and $B$ is a Borel set.