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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.

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2 Answers 2

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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.

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  • $\begingroup$ How do I know there exist such set $B_k$ and that it's measure is exactly this sum? I sense it got something to do with the fact that outer measure of A is finite- $\endgroup$
    – Meow
    Dec 18, 2013 at 16:15
  • $\begingroup$ $\lambda^{\ast}(A)$ is the infimum of a certain set $S$; so for any $\epsilon > 0$, $\lambda^{\ast}(A) + \epsilon$ cannot be an upper bound for $S$, and so there is an $s \in S$ such that $$s > \lambda^{\ast}(A) + \epsilon $$ $\endgroup$ Dec 18, 2013 at 16:18
  • $\begingroup$ I have one more question =) In definition of Lebesgue outer measure (from my question) do we assume that intervals $(a_i,b_i)$ are disjoint which would imply $\lambda^*(\bigcup_i(a_i,b_i))=\sum_i(b_i-a_i)$ $\endgroup$
    – Meow
    Dec 18, 2013 at 17:15
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    $\begingroup$ If a pair of open intervals are not disjoint you can always replace them by their union. This applies to infinite collections as well. $\endgroup$ Dec 18, 2013 at 17:27
  • $\begingroup$ math.stackexchange.com/questions/3891592/… hello, could you please clarify this answer? @PrahladVaidyanathan $\endgroup$ Nov 2, 2020 at 23:20
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Take a sequence of open sets $B_n$ as in your definition whose measure tends to the infimum $\lambda^\ast(A)$, and set $B=\cap_n B_n$.

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