Suppose $\{f_t\}$ is an uncountable family of measurable functions , then can we show that $F(x):=\sup_t f_t(x)$ measurable?

Question

Suppose $\{f_t\}_{t\in R }$ is a family of measurable functions , then can we show that $F(x):=\sup_t f_t(x)$ measurable?

Recall that when $\{f_n\}$ is a countable sequence of measurable functions , then $F:=\sup_n f_n$ is measurable , since $$F^{-1}(a,\infty]=\bigcup_1^{\infty}f_n^{-1}(a,\infty]$$ And a countable union of measurable sets is measurable . However , uncountable union of measurable sets need not to be measurable (every unmeasurable set can be written as uncountable union of single point). So we can not proceed as before .
I think the answer to the above question is Yes , at least for $f_t \ge 0$ since we have the following well known theorem for maximal operators:

However , I can not show why $F$ is measurable . Any help would be very appreciate .

Answer 1

Let $A \subseteq \mathbb{R}$ be non-measurable. Then $A$ is uncountable, but the indicator for $A$ can be written as a pointwise supremum of indicators of singletons (i.e. the points in $A$).