# Pointwise Supremum of Measurable Functions is Measurable

## Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space, and let $I$ be a countable set.

Let $\left({f_i}\right)_{i \in I}$, $f_i: X \to \overline \R$ be an $I$-indexed collection of $\Sigma$-measurable functions.

Then the pointwise supremum $\displaystyle \sup_{i \mathop \in I} f_i: X \to \overline \R$ is also $\Sigma$-measurable.

## Proof

Let $a \in \R$; for all $i \in I$, we have by Characterization of Measurable Functions that:

$\left\{{f_i > a}\right\} \in \Sigma$

and as $\Sigma$ is a $\sigma$-algebra and $I$ is countable, also:

$\displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \in \Sigma$

We will now show that:

$\displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\} = \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\}$

First, observe that for all $i \in I$:

$f_i \left({x}\right) \le \displaystyle \sup_{i \mathop \in I} f_i \left({x}\right)$

and hence:

$\left\{{f_i > a}\right\} \subseteq \displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$
$\displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \subseteq \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$

Next, suppose that:

$x \notin \displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\}$

Then, by definition of union:

$\forall i \in I: f_i \left({x}\right) \le a$

which is to say that $a$ is an upper bound for the $f_i \left({x}\right)$.

Hence, by definition of supremum, it follows that:

$\displaystyle \sup_{i \mathop \in I} f_i \left({x}\right) \le a$

and therefore:

$x \notin \displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$

Thus, we have shown:

$\displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\} = \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \in \Sigma$

and by Characterization of Measurable Functions, it follows that $\displaystyle \sup_{i \mathop \in I} f_i$ is measurable.

$\blacksquare$