# On the relation of a certain set function with the outer measure

Let $A$ be a set and define $m^{**}(A) \in [0,\infty]$ by:

$m^{**} (A)= \inf\{m^{*}(\mathcal{O}) \ | \ A \subset \mathcal{O}, \mathcal{O} \ \text{open} \}$

How is $m^{**}$ related to the outer measure $m^{*}$?

I would appreciate any help on this!!

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## 1 Answer

Here's an idea for the proof when $m$ is Lebesgue measure on $(0,1)$. A: $m^{**} = m^*$

It is possible to define $m^*(A)$ as the infimum of all quantities $$\sum_{i = 1}^{\infty} (b_i - a_i)$$ where $0< a_i < b_i < 1$ and $A \subset \cup_{i = 1}^{\infty} (a_i, b_i)$. Now assume we've chosen our intervals so that $\sum_1^{\infty} b_i - a_i \leq m^*A + \epsilon$. Clearly $U = \cup_1^{\infty} (a_i, b_i)$ is open and covers $A$, so we obtain $m^{*}U = m^* \cup_1^{\infty} (b_i - a_i) \leq \sum_{1}^{\infty}(b_i - a_i)$ where we've used the countable subadditivity of outer measure, as well as $m^*(a_i, b_i) = b_i - a_i$. We deduce that $$m^{**}A \leq m^*U \leq m^*A + \epsilon$$ from which $m^{**}A \leq m^* A$ follows by taking $\epsilon \rightarrow 0$.

For the other direction, choose $U \supset A$ open such that $m^*U \leq m^{**}A + \epsilon$. Then $U$ is the countable union of disjoint open intervals $\cup_{i = 1}^{\infty}(a_i, b_i)$.

From here, if you already know that outer measure restricted to the Borel sets is genuine Lebesgue measure, then our work is done, since $$m^*A \leq \sum_{i = 1}^{\infty} b_i - a_i = mU = m^*U \leq m^{**}A + \epsilon$$ Now take $\epsilon \rightarrow 0$.

The proof easily generalizes to Lebesgue Stieltjes measure on the real line(the primary fact used, that an open set is the countable disjoint union of intervals, holds on the real line), but there's a few issues to deal with when $A$ is unbounded.

For more general topologies, I would worry for instance that my 'primary fact' is now false; for e.g. there's the Long Line, for which this property most certainly does not hold. However if the space is "small" enough, I do think it's still true.

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