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Let $(X,\mathcal{X})$ be a topological space and $\mu:\mathcal{B}(\mathcal{X}) \to [0,\infty]$ a measure on $(X,\mathcal{B}(\mathcal{X}))$ the Borelian $\sigma$-algebra on $(X,\mathcal{X})$.

The outer measure associated to $\mu$ is defined for any set $A\subset X$ by $$ \mu^*(A) = \inf \left\{ \sum_i \mu(A_i), A_i \in \mathcal{B}(\mathcal{X}) \mbox{ and } A \subset \bigcup_i A_i \right\}.$$

An outer measure $\nu$ is said to be Borel regular if the Borel sets are measurables (in the sense that for a Borel set $B$ and for $A\subset X$, $\nu(A)=\nu(A\cap B) + \nu(B\backslash A)$) and if for any $A\subset X$, there exists a Borel set $B$ containing $A$ with $\nu(A)=\nu(B)$.

Is $\mu^*$ always Borel regular? If not, what if $X$ is a metric space?

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Ok actually this was really straightforward. Let $A \subset X$. If $\mu^*(A) = \infty$, the result holds. Otherwise, for $k>0$, let $(A_i^k)_i$ be a cover of $A$ with Borel sets such that $$\mu\left(\bigcup_i A_i^k\right) \leq \sum_i \mu(A_i^k) \leq \mu^*(A) + \frac{1}{k}.$$ One can always assume that $\bigcup_i A_i^k \subset \bigcup_i A_i^{k-1}$. Therefore, setting $B=\bigcap_k \bigcup_i A_i^k$, $$\mu(B)=\lim_k \mu\left(\bigcup_i A_i^k\right) \leq \mu^*(A)\leq \mu^*(B)=\mu(B).$$ As $B$ is a Borel set, the conclusion holds.

Remark: the motivation for this question was the fact that there are a priori two different definitions of Radon measures found in the litterature: either (i) a Radon measure is an outer measure which is Borel regular and finite on compact sets, or (ii) it is a measure on the Borelian $\sigma$-algebra which is finite on compact sets. This lemma proves that the two notions coincides: if $\mu$ is Radon(ii), then $\mu^*$ is Radon(i) with $\mu^* \equiv \mu$ on the Borel sets.

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