Suppose $\mu$ is a finitely additive measure on $X$ (aka “content”) with $\mu(X) < \infty$, defined on an algebra of sets $\mathcal A$. Let
$$\mu^\*(Y) = \inf \{ \mu(E) : E \in \mathcal A \wedge E \supseteq Y \}.$$
Question: Is it true that for all $Y \subseteq X$, there is an extension $\nu$ of $\mu$, where $\nu$ is a content defined at $Y$, such that $\nu(Y) = \mu^\*(Y)$? If so, can this be proven without the axiom of choice?



Extending contents
I believe part of the question is answered in the following paper:


[Łoś, J.; Marczewski, E. Extensions of measure. Fund. Math. 36 (1949), 267–276.](https://www.impan.pl/pl/wydawnictwa/czasopisma-i-serie-wydawnicze/fundamenta-mathematicae/all/36/0/93726/extensions-of-measure)


Let $c$ be any number between (inclusive) the inner and outer measure of $Y$. Then there is an extension $\nu$ which assigns $c$ to $Y$, with the domain of $\nu$ being the algebra of sets generated by the set $Y$ together with the sets in the domain of $\mu$.
($\mu$ is the measure to be extended.)


I am not certain about the role (if any) of the axiom choice.


Roger Purves


