Let $\mathcal F$ be a field of subsets of a set $\Omega$. Equip the space $[0,1]^\mathcal F$ of functions from $\mathcal F$ into $[0,1]$ with the product topology. Then, the set $\Delta$ of finitely additive probability measures on $\mathcal F$ is a convex and compact subset of $[0,1]^\mathcal F$.

If $\mu \in \Delta$, define the inner measure $\mu^i: 2^\Omega \to [0,1]$ for $\mu$ by $$\mu^i(A) = \sup \big\{\mu(F): F \subset A, F \in \mathcal F \big\}, \ A \subset \Omega.$$ We can view the inner measure as a mapping $\mu \mapsto \mu^i$ from $\Delta$ into $[0,1]^{2^\Omega}$.

Question. Is the inner measure mapping continuous? That is, if $\mu_{\alpha}$ is a net in $\Delta$ that converges to $\mu$ (i.e. $\mu_\alpha(F) \to \mu(F)$ for all $F \in \mathcal F$), then is it true that $\mu^i_\alpha \to \mu^i$ (i.e. $\mu_\alpha^i(A) \to \mu^i(A)$ for every subset $A$ of $\Omega$)?

A similar question arises for outer measure, though I assume the answers are the same.


Let $\Omega = \mathbb{N}$ and let $\mathcal{F}$ be the field consisting of all finite and cofinite subsets of $\mathbb{N}$. Let $\mu_n = \delta_{2n}$ be a point mass at the integer $2n$, and let $\mu$ be the finitely additive measure that assigns measure $0$ to every finite set and $1$ to every cofinite set. Then $\mu_n(F) \to \mu(F)$ for every $F \in \mathcal{F}$.

Let $A \subset \mathbb{N}$ be the set of all even integers. Then we have $\mu_n^i(A) = 1$ for every $n$ but $\mu^i(A) = 0$.

  • $\begingroup$ Thanks. On the other hand, if $\mathcal F$ is closed under arbitrary unions, then I think the result holds. I wonder if something like that is necessary, though. $\endgroup$ – aduh Mar 29 '20 at 23:30

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