Good day, remembering the definition: The function $f(x)$ defined on the set $E$ is said to be measurable if the bounded set $E$ is measurable and if the set $E\cap \left \{ x:f(x)> a \right \}$ is measurable for all $a$.

Prove that the least upper bound of a finite or denumerable set of measurable functions is a measurable function.

Can I use this result? If $f(x)$ is a measurable function defined on the set $E$, then the sets $E\cap \left \{ x:f(x) \geq a \right \}$, $E\cap \left \{ x:f(x)= a \right \}$, $E\cap \left \{ x:f(x) \leq a \right \}$ are measurables for all $a$.

  • $\begingroup$ Well, of course you can. The first and the last conditions are in fact both equivalent to the definition of measurable (real-valued or extended-real-valued) function. $\endgroup$
    – user228113
    Sep 26, 2017 at 23:41

1 Answer 1


You need only show that one of them is measurable for every $a$. To do this, note that if $f$ is the supremum of a sequence $(f_n)$ of functions, then $f(x)>a$ if and only if $f_n(x)>a$ for some $n$. Therefore, if each $f_n$ is measurable, then $$ \{x : f(x) > a\} = \bigcup_{n=1}^n\{x : f_n(x)>a\}, $$ which shows that $$ E\cap \{x : f(x) > a\} = E\cap\left(\bigcup_{n=1}^n\{x : f_n(x)>a\}\right) =\bigcup_{n=1}^n(E\cap \{x : f_n(x)>a\}) $$ is a countable union of measurable sets.

It's important that the supremum is taken over countably many functions. To see why, let $N\subseteq[0,1]$ be a Lebesgue nonmeasurable set. Then the characteristic function $1_N$ is not Lebesgue measurable, but the family $\{1_{\{x\}} : x\in N\}$ consists of measurable functions and $1_N$ is their supremum.


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