Measurability of supremum of measurable functions

Question

It is known that the supremum of a sequence of measurable functions is measurable. This fact is easily proved by considering the inverse images of sets of the form: $(c,\infty)$, for example, here. Let $f_t:X \to \mathbb{R}$ - measurable functions on $(X,\mathbb{F},P) - $ probability space. I want to prove that, $\sup\limits_{t \in \mathbb{R}}$ $f_t(x)$ is also measurable.

Answer 1

There are nice links in the comments and similar questions on related pages of this site, mostly with negative answers to the generally cast question, but I thought I would add a positive answer to an important special case. If the functions $f_t(x)$ have continuity in the $t$-parameter for each fixed $x$, then $\sup_t f_t(x)$ is measurable.

Proposition. Suppose $T\subset\mathbb R$ is an arbitrary set, and that for each $t\in T$ we are given $f_t(x)$ a measurable function (of $x$). If for each fixed $x$, the function $t\mapsto f_t(x)$ is continuous, then $$\sup_{t\in T}f_t(x)\ \text{is measurable.}$$ Proof. Measurability of $\sup_{t\in T}f_t$ is equivalent to being able to recognize the level sets $\{x:\sup_{t\in T}f_t(x)>c\}$ as measurable sets. By definition of the supremum, $$ \{x:\sup_{t\in T}f_t(x)>c\} = \bigcup_{t\in T}\{x:f_t(x)>c\}. $$ Let $J\subset T$ be a countable dense set. By continuity in $t$, if $f_t(x)>c$ for some $t\in T$, then for $t'\in J$ close enough to $t$, we also have $f_{t'}(x)>c$. Therefore, we can also write $$ \bigcup_{t\in T}\{x:f_t(x)>c\} = \bigcup_{t\in J}\{x:f_t(x)>c\}, $$ the latter set being measurable because it is a countable union of measurable sets. $\square$

This generalizes in a straightforward way to index sets $T$ with a suitable topology (they should possess countable dense subsets) where $t\mapsto f_t(x)$ is assumed continuous.