# Find an example about the supremum of an uncountable family of real-valued measurable functions need not be measurable

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

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.

• So if $N$ is countable, then we can let $N\subset \cup_{x\in N}\{x\}$ to show that $N$ is measurable? – user469065 Sep 26 '19 at 0:54
• We need equality : $N =\cup_{ x \in N}\{x\}$, then if $N$ is countable it is measurable. – Teresa Lisbon Sep 26 '19 at 1:00
• @actoh Follow your idea, I write answer. Could you see is it correct? – user469065 Sep 26 '19 at 1:10
• 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 $\mathbb 1_N$. – Teresa Lisbon Sep 26 '19 at 1:13