Measurability of the infimum of a class of functions

Ref:

[1] Supremum of a class of functions

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) <a\} = \cup_{y\in Y} \{x: g_{y} (x)< a\}.

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} <a\},

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


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