Let $\Omega_1$ and $\Omega_2$ be two disjoint measurable sets in $\Bbb R$, and then $\int_{{\Omega _1} \cup {\Omega _2}} f = \int_{{\Omega _1}} f + \int_{{\Omega _2}} f $ where $f$ is a measurable function. We call this as "set extension".

Let $\Omega_i,i=1,2,...$ be countable many disjoint sets, then $\int_{\bigcup\limits_{i = 1}^\infty {{\Omega _i}} } f = \sum\limits_{i = 1}^\infty {\int_{{\Omega _i}} f } $, which can be proved by dominated convergence theorem. We say this is a countable set extension.

My question is does similar equation holds for uncountable set extension? That is, let $\Omega_i, i\in I$ be a family of disjoint sets where $I$ is an uncountable index set, does it hold that $\int_{\bigcup\nolimits_{i \in I} {{\Omega _i}} } f = \sum\nolimits_{i \in I} {\int_{{\Omega _i}} f } $, and how can we prove this? Thank you!

  • $\begingroup$ $f$ needs to be more than measurable $\endgroup$
    – zhw.
    Dec 10 '15 at 19:03

No, with one reason being that every subset of $\mathbb{R}$ can be written as a (possibly uncountable) union of singleton sets. If we take $f$ to be, say, constant $1$ and our family to be $\{x\}_{x \in \mathbb{R}}$ (here just using the standard Lebesgue measure on $\mathbb{R}$), then $$\infty = \int_{\bigcup_{x \in \mathbb{R}} \{x\}} 1 \ne \sum_{x \in \mathbb{R}} \int_{\{x\}} 1 = \sum_{x \in \mathbb{R}} 0 = 0$$

Two other issues come up:

  • The example above doesn't even have any technical issues regarding convergence or existence of sums over uncountable sets. For example, if $f$ is a positive function, then the sum converges only when $I$ is at most countable.

  • Measurability does not interact nicely with uncountable unions. Adapting the example above to use a non-measurable set rather than $\mathbb{R}$, then the right side may make sense while the left does not.


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