Measurability of the infimum of a class of functions

Ref:

In the earlier post [1], we discussed measurability of the infimum of a class of measurable functions. In particular, for the infimum of a class of measurable functions as a function, we can show that it may not be measurable. Therefore, we shall need additional conditions to have the infimum function measurable. In this post, we show that the infimum function is

• measurable if the class size is countable;
• lower semicontinuous (thus measurable) if each function in the class is lower semicontinuous.

For simplicity, we consider the real number space ${\mathbb R}$, but all discussions below extend to a complete metric space. One can use all open balls to generate the smallest ${\sigma}$-algebra, which is referred as Borel ${\sigma}$-algebra, denoted by ${\mathcal B}$. A set in ${\mathcal B}$ is called a Borel set.

Given ${g:\mathbb R^{2} \mapsto \mathbb R}$, the infimum function ${f}$ is defined by, for some set ${Y}$

$\displaystyle f(x) = \inf_{y \in Y} g(x, y).$

Our question is when ${f}$ is (Borel) measurable? Recall ${f}$ is measurable, if ${f\in \mathcal B/\mathcal B}$, i.e.

$\displaystyle f^{-1} (B) \in \mathcal B, \ \forall B\in \mathcal B.$

In general,

• even if ${g}$ is jointly measurable, i.e. ${g\in \mathcal B(\mathcal R^{2}) / \mathcal B}$, the infimum ${f}$ may not be measurable.

Indeed, we can consider the following example: Let ${g(x, y) = I_{y}(x)}$ and ${Y\notin \mathcal B}$. Then ${g}$ is measurable, since ${\{g = 1\}= \{x = y\}}$. But ${f}$ is not measurable, since ${\{f = 1\} = Y\notin \mathcal B}$.

However,

• If ${g_{y}(\cdot) := g(\cdot, y)}$ is continuous for each ${y}$, then ${f}$ is upper semicontinuous, hence is measurable.

To see that, we write for any ${a\in \mathbb R}$

$\displaystyle \{x: f(x)< a\} = \{x: \inf_{y \in Y} g(x, y)

Union of possibly uncountably many open sets is open, and ${\{x: f(x)< a\}}$ is open. So ${f}$ is USC.

Alternatively,

• If ${g}$ is measurable and ${Y}$ is a countable set, then ${f}$ is measurable.

This claim follows from

$\displaystyle \{f< a\} = \cup_{y\in Y} \{g_{y}

i.e. Borel ${\sigma}$-algebra is closed under countable union.