Pointwise Minimum of Measurable Functions is Measurable

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions.


Then the pointwise minimum $\min \left({f, g}\right): X \to \overline{\R}$ is also $\Sigma$-measurable.


Proof

For all $x \in X$ and $a \in \R$, we have by Min yields Infimum of Parameters that:

$a \le \min \left({f \left({x}\right), g \left({x}\right)}\right)$

iff both $a \le f \left({x}\right)$ and $a \le g \left({x}\right)$.


That is, for all $a \in \R$:

$\left\{{x \in X: \min \left({f \left({x}\right), g \left({x}\right)}\right) \ge a}\right\} = \left\{{x \in X: f \left({x}\right) \ge a}\right\} \cap \left\{{x \in X: g \left({x}\right) \ge a}\right\}$

By Characterization of Measurable Functions: $(1) \implies (4)$, the two sets on the RHS are elements of $\Sigma$, i.e. measurable.


By Sigma-Algebra Closed under Intersection, it follows that:

$\left\{{x \in X: \min \left({f \left({x}\right), g \left({x}\right)}\right) \ge a}\right\} \in \Sigma$

Hence $\min \left({f, g}\right)$ is measurable, by Characterization of Measurable Functions: $(4) \implies (1)$.

$\blacksquare$


Sources