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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 $f(x)\geq 0=\mathbb{1}_{e}$.

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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.

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  • $\begingroup$ So if $N$ is countable, then we can let $N\subset \cup_{x\in N}\{x\}$ to show that $N$ is measurable? $\endgroup$ – user469065 Sep 26 '19 at 0:54
  • $\begingroup$ We need equality : $N =\cup_{ x \in N}\{x\}$, then if $N$ is countable it is measurable. $\endgroup$ – Teresa Lisbon Sep 26 '19 at 1:00
  • $\begingroup$ @actoh Follow your idea, I write answer. Could you see is it correct? $\endgroup$ – user469065 Sep 26 '19 at 1:10
  • $\begingroup$ It is correct, but you went over the supremum part in way too much detail! It is very simple : indeed, every $f \in S$ takes only values $0$ and $1$, and at each point in $N$ there is a function which takes the value $1$, so the pointwise supremum is $1$ which means the supremum is $\mathbb 1_N$. $\endgroup$ – Teresa Lisbon Sep 26 '19 at 1:13

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