3
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ $f$ needs to be more than measurable $\endgroup$
    – zhw.
    Dec 10 '15 at 19:03
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.