# Pointwise Infimum 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 infimum $\displaystyle \inf_{i \mathop \in I} f_i: X \to \overline{\R}$ is also $\Sigma$-measurable.

## Proof

From Infimum as Supremum, we have the Equality of Mappings:

$\displaystyle \inf_{i \mathop \in I} f_i = - \left({\sup_{i \mathop \in I} \, \left({- f_i}\right)}\right)$

Now, from Negative of Measurable Function is Measurable and Pointwise Supremum of Measurable Functions is Measurable, it follows that:

$\displaystyle - \left({\sup_{i \mathop \in I} \, \left({- f_i}\right)}\right)$

is a measurable function.

Hence the result.

$\blacksquare$