Extending contents

Question

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?

Answer 1

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.

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

Answer 2

$\let\bez\smallsetminus\let\sdif\vartriangle$Such as extension exists, and it can be constructed explicitly without using the axiom of choice.

Let $\mathcal B$ denote the algebra generated by $\mathcal A\cup\{Y\}$, i.e., $$\mathcal B=\bigl\{(Z_0\cap Y)\cup(Z_1\bez Y):Z_0,Z_1\in\mathcal A\bigr\}.$$ Put $$\mathcal F=\{E\in\mathcal A:E\supseteq Y\}.$$ Note that $\mathcal F$ is a filter of $\mathcal A$. For any $Z\in\mathcal A$, we define $$\begin{align} \mu_0(Z)&=\inf\{\mu(Z\cap E):E\in\mathcal F\},\\ \mu_1(Z)&=\sup\{\mu(Z\bez E):E\in\mathcal F\}. \end{align}$$ Since $\mu$ is monotone and $\mathcal F$ is a filter, we have $$\mu_0(Z)=\inf\{\mu(Z\cap E):E\in\mathcal F,E\subseteq E_0\}$$ for any $E_0\in\mathcal F$. In particular, $$Z\cap Y=Z'\cap Y\implies\mu_0(Z)=\mu_0(Z'),\tag1$$ because $E_0=X\bez(Z\sdif Z')\in\mathcal F$, and $Z\cap E=Z'\cap E$ for any $E\subseteq E_0$. Also, $$Z\bez Y=Z'\bez Y\implies\mu_1(Z)=\mu_1(Z')\tag2$$ as then $Z\bez E=Z'\bez E$ for all $E\in\mathcal F$. Thus, we can define $\nu\colon\mathcal B\to\mathbb R_{\ge0}$ by $$\nu\bigl((Z_0\cap Y)\cup(Z_1\bez Y)\bigr)=\mu_0(Z_0)+\mu_1(Z_1).\tag3$$ This is independent of the choice of $Z_0$ and $Z_1$ by (1) and (2).

Clearly, $$\nu(Y)=\mu_0(X)+\mu_1(\varnothing)=\mu^*(Y).\tag4$$ For any $Z\in\mathcal A$, we have $$\mu_1(Z)=\sup\bigl\{\mu(Z)-\mu(Z\cap E):E\in\mathcal F\bigr\}=\mu(Z)-\mu_0(Z),$$ thus $$\nu(Z)=\mu(Z).\tag5$$ It remains to prove that $\nu$ is additive, which follows from the properties $$\begin{align} Z\cap Z'\cap Y=\varnothing&\implies\mu_0(Z\cup Z')=\mu_0(Z)+\mu_0(Z'),\tag6\\ Z\cap Z'\subseteq Y&\implies\mu_1(Z\cup Z')=\mu_1(Z)+\mu_1(Z').\tag7 \end{align}$$ For (6), put $E_0=X\bez(Z\cap Z')\in\mathcal F$; we have $$\begin{align} \mu_0(Z)+\mu_0(Z')&=\inf\{\mu(Z\cap E):E\in\mathcal F\}+\inf\{\mu(Z'\cap E):E\in\mathcal F\}\\ &=\inf\{\mu(Z\cap E)+\mu(Z'\cap E'):E,E'\in\mathcal F\}\\ &=\inf\{\mu(Z\cap E)+\mu(Z'\cap E):E\in\mathcal F,E\subseteq E_0\}\\ &=\inf\{\mu((Z\cup Z')\cap E):E\in\mathcal F,E\subseteq E_0\}=\mu_0(Z\cup Z'), \end{align}$$ where the third equality follows from $\mathcal F$ being a filter and the monotonicity of $\mu$.

The equation (7) is proved in the same way, except we do not have to bother with $E_0$.

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We have constructed a content $\nu\supseteq\mu$ such that $$\nu(Y)=\mu^*(Y).$$ Using the same construction with $X\bez Y$ in place of $Y$, we obtain a content $\nu'\supseteq\mu$ such that $$\nu'(Y)=\mu_*(Y)=\sup\{\mu(E):E\in\mathcal A,E\subseteq Y\}.$$ Then, for any $y$ such that $\mu_*(Y)\le y\le\mu^*(Y)$, $$t\nu(Y)+(1-t)\nu'(Y)$$ is a content that gives $Y$ value $y$, where $t=(y-\mu_*(Y))/(\mu^*(Y)-\mu_*(Y))\in[0,1]$.