TITLE: Find an example about the supremum of an uncountable family of real-valued measurable functions need not be measurable
QUESTION [6 upvotes]: Show that by way of an explicit example that the supremum of an uncountable family of real-valued measurable functions need not be measurable. 

Is there such an example?
My solution:
Consider Lebesgue measure on $\mathbb{R}$. If we assume that set $E$ is a non-measurable set(such as Vitali set which is not Lebesgue measurable), then consider the collection of all indicator function of $e\in E$, that is, A:= $\{1_{e}: e\in E\}$ which is uncountable. In fact, if $E$ is not uncountable, then at most countable set $E$ is measurable. Notice that the the supremum of $A$ is $\mathbb{1}_{E}$. Indeed, if $\forall x\in E$, then $\mathbb{1}_{E}(x)=1\geq \mathbb{1}_{e}(x)$. If $\forall x\notin E$, then $\mathbb{1}_{E}(x)=0=\mathbb{1}_{e}(x)$. Also, we claim that if there exists function $f$ such that $f<\mathbb{1}_{E}$, then $f<\mathbb{1}_{e}$ which means $\mathbb{1}_{E}$ is the supremum of $A$. Indeed, if $\forall x\in E$, then $f(x)<\mathbb{1}_{E}(x)=1$ which implies $f(x)<\mathbb{1}_{x}(x)$. Also, If $\forall x\notin E$, then $f(x)<\mathbb{1}_{E}(x)=0$ which implies $f(x)\geq 0=\mathbb{1}_{e}$.

REPLY [6 votes]: Work over the Lebesgue measure on $\mathbb R$. We use existence of a non-measurable set, and the fact that every singleton is measurable.
Let $N$ be a non-measurable set. Then $\mathbb 1_N$, the indicator function of $N$, is not measurable, since $\{\mathbb 1_N \geq 1\} = N$ is not measurable.
Now, take the set of all singletons of $N$, and their indicators. So you have :
$$
S = \{\mathbb 1_{\{x\}} : x \in N\}
$$

Show that $S$ consists of real valued measurable functions.
Show that the supremum $\sup_{f \in S} f = \mathbb 1_N$, so it is not measurable.

Note that $N$ is uncountable, since any at most countable set is measurable.