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Let $(X_\lambda,B_\lambda,\mu_\lambda)$ be a disjoint collection of measure spaces indexed by $\lambda$. Define $X$ to be the union of all the $X_\lambda$. Define $B=\{S \subseteq X: S \cap X_\lambda \in B_ \lambda\ \forall \lambda\}$. Define $\mu(E)=\sum_\lambda \mu(E \cap X_\lambda)$.

Now, assume that $(X,B, \mu)$ is a $\sigma$-finite measure space. I want to show that all but countably many $\mu_\lambda(X_\lambda)$ have to be $0$.

I figured out that each $\mu_\lambda$ has to be $\sigma$-finite. This is because, writing $X=A_1 \cup A_2 \cup...$ as a union of finite measure sets, each $\mu_\lambda(A_i \cap X_ \lambda) \leq \mu(A_i)$ and hence must be finite. But I don't think same trick will work for proving the above.

Pascal's Wager
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1 Answers1

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This really comes down to this problem: The sum of an uncountable number of positive numbers. Using this we can make the following argument.

Suppose $X=\sqcup_{n\geq 1} A_n$ with $\mu(A_n)<\infty$ for each $n\geq 1$. Then, for each $n\geq 1$, we have $\sum_\lambda \mu_\lambda(A_n\cap X_\lambda)<\infty$ so the above problem implies that $\mu_\lambda(A_n\cap X_\lambda)=0$ for all but at most countably many $\lambda$. Hence, for each of these all but at most countably many such $\lambda$, we have $$\begin{align}0=\sum_{n\geq 1} \mu_\lambda(A_n\cap X_\lambda)&=\mu_\lambda\left(\sqcup_{n\geq 1} (A_n\cap X_\lambda)\right)\\&=\mu_\lambda((\sqcup_{n\geq 1} A_n)\cap X_\lambda)=\mu_\lambda(X\cap X_\lambda)=\mu_\lambda(X_\lambda).\end{align}$$

Dave
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