Definition of measurable set: A set $E$ measurable if $$m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)$$ for every subset of $A$ of $\mathbb R^n$.

Definition of Lebesgue measurable function: Given a function $f: D \to \mathbb R ∪ \{+\infty, -\infty\}$, defined on some domain $D \subset \mathbb{R}^n$, we say that $f$ is Lebesgue measurable if $D$ is measurable and if, for each $a\in[-\infty, +\infty]$, the set $\{x\in D \mid f(x) > a\}$ is measurable.

So if $f$ is a extended real valued function defined on $\mathbb R^n$ ($f: \mathbb R^n \to [-\infty, +\infty]$) and f is Lebesgue measurable on each set $E_k \subset \mathbb R^n, k \in \mathbb N^+$, will it be Lebesgue measurable on $\bigcup_{k=1}^{+\infty} E_k$ as well?

I think for finite positive integer $n$, $f$ will be Yes on $\bigcup_{k=1}^{n} E_k$ because $\{x \in E_1 \cup E_2: f(x) > t \} = \{x \in E_1: f(x) > t \} \cup \{x \in E_2: f(x) > t \}$. But I'm not sure for countable union of them(for example, countable union of closed sets can be a open set).

  • $\begingroup$ $\bigcap$ is conventionally used in expressions like $\displaystyle\bigcap_{k=1}^n A_k$ and $\cap$ is used in things like $A\cap B$ or $A_1\cap\cdots\cap A_n$. I edited accordingly. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 19 '15 at 3:49
  • $\begingroup$ @MichaelHardy: Thanks. I got it. $\endgroup$ – Bear and bunny Jul 19 '15 at 4:06


Assume for now the $E_k$ are disjoint. For each $k$ define $f_k = f \chi_{E_k}$. Then $f_k = f$ on $E_k$ and $0$ off $E_k$ so $f_k$ is measurable. For each $x \in \bigcup E_k$ you have $$f(x) = \sum_k f_k(x)$$ so that $f$ is a sum of measurable functions, hence measurable.

If the $E_k$ aren't disjoint, you can find measurable sets $G_k \subset E_k$ with the property that the $G_k$ are disjoint and $\bigcup E_k = \bigcup G_k$. Proceed as above.

  • $\begingroup$ $\bigcup E_k = \mathbb R^n$, the domain where $f$ defined? $\endgroup$ – Bear and bunny Jul 19 '15 at 2:35
  • $\begingroup$ That isn't necessary. The countable union of the $E_k$ is measurable, and you are given no information about $f$ off that set. $\endgroup$ – Umberto P. Jul 19 '15 at 19:42

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