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Let $X$ be a measure space, $\mu$ its measure function. Suppose $X$ is a disjoint union of a family of measurable sets $\{X_\alpha : \alpha\in A\}$. Suppose $\mu(X_\alpha)\gt 0$ for all $\alpha\in A$. If $A$ is countable, it is quite possible that $\mu(X) \lt \infty$. For example, this is the case if $A =\{1,2,\dots\}$ and $\mu(X_\alpha) = 1/2^\alpha$ for each $\alpha \in A$. Now suppose $A$ is uncountable. Is the following assertion true?

$\mu(X) = \infty$.

This seems true, but I was unable to prove it.

Makoto Kato
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2 Answers2

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Write

$$X=\bigcup_{n=1}^\infty Y_n$$

where $Y_n=\bigcup \{ X_\alpha : \alpha \in A \text{ and } \mu(X_\alpha)>1/n \}$. Clearly $Y_n \subseteq X$ for every $n$. If $A$ is uncountable, then at least one of the $Y_n$, say $Y_N$, contains infinitely many $X_\alpha$. By considering countably many such $X_\alpha$, say $X^{(1)},X^{(2)},\dots$, we conclude that

$$\mu(X) \geq \mu(Y_N)=\sum_{i=1}^\infty \mu(X^{(i)}) \geq \sum_{i=1}^\infty 1/N = +\infty.$$

Informally speaking, "any uncountable sum of positive terms is infinite".

(As an aside, I don't think this proof is optimal with respect to the strength of the choice axiom used. There is no way to get around proving "a countable union of finite sets is countable", but there is probably a way to work around having to extract a countable set from an unknown, possibly uncountable set.)

Ian
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yes it is true. since the sets $\{X_{\alpha}|\alpha\in A\}$ are disjoint we have: ( for proof, see https://math.stackexchange.com/a/4666886/1034911)

$\mu(\underset{\alpha\in A}{\bigcup}X_{\alpha})\geq\underset{\alpha\in A}{\sum}\mu(X_{\alpha})$

where

$\underset{\alpha\in A}{\sum}\mu(X_{\alpha})=\underset{\underset{\text{F is finite}}{F\subseteq A}}{\sup}\Bigg(\underset{\alpha\in F}{\sum}\mu(X_{\alpha})\Bigg)$

and we know that if $\mu(X_{\alpha})>0$ for every $\alpha\in A$, then we have $\underset{\alpha\in A}{\sum}\mu(X_{\alpha})=\infty$ ( for proof, see The sum of an uncountable number of positive numbers).

As a result, the relation $\mu(X)=\mu(\underset{\alpha\in A}{\bigcup}X_{\alpha})\geq \underset{\alpha\in A}{\sum}\mu(X_{\alpha})= \infty$ must hold.

Amirhossein Haddadian
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