Let us call a function $\mu : \mathcal{H} \to [0, \infty]$ $\sigma$-subadditive, if $\mu(A) \le \sum_{i \in I} \mu(A_i)$ for every $A \in \mathcal{H}$ and for every countable family $(A_i)_{i \in I}$ in $\mathcal{H}$ such that $A \subseteq \bigcup_{i \in I} A_i$. For example every outer measure is $\sigma$-subadditive.

Now let $\mu_i : \mathcal{H}_i \to [0, \infty]$ be $\sigma$-subadditive functions for $i \in \{1,2\}$. Let $\mathcal{H} = \{A_1 \times A_2 ; \, A_i \in \mathcal{H}_i\}$, and $\mu \colon \mathcal{H} \to [0, \infty]$, $A_1 \times A_2 \mapsto \mu_1(A_1) \mu_2(A_2)$, where $A_1 \times A_2$ denotes the Cartesian product of the sets $A_1$ and $A_2$. (This function is well defined, because if $A_1 \times A_2$ is empty, then $\mu_1(A_1) = 0$ or $\mu_1(A_2)=0$. We use the convention $0 \cdot \infty = \infty \cdot 0 = 0$.)

Is $\mu$ also $\sigma$-subadditive?

Remark: We could replace $\mu_i$ by the outer measure generated by $\mu_i$, so in fact we may assume that $\mu_1$ and $\mu_2$ are outer measures on some sets $X_1$ and $X_2$.

  • $\begingroup$ Is this homework? $\endgroup$ – Nik Weaver Mar 17 '14 at 18:27
  • $\begingroup$ No, it's not homework. Why, is it so easy? $\endgroup$ – user42355 Mar 17 '14 at 18:51
  • $\begingroup$ I don't know whether it's easy until I try it. It kind of looks like homework, though. What is the context? (And what is ${\cal H}$?) $\endgroup$ – Nik Weaver Mar 17 '14 at 20:27
  • $\begingroup$ I was looking at different versions of Fubini's and Tonelli's theorems. For these one needs a basic understanding of product measures. This is just a simple question in that area. My conjecture is that $\mu$ is not always $\sigma$-subadditive, mainly because otherwise I would have seen this statement before. The smallest possible counterexample could occur when $|X_1| = |X_2| = 3$. However I think in that case $\mu$ is $\sigma$-subadditive, but this is not so easy to check. $\endgroup$ – user42355 Mar 17 '14 at 21:36

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