# A collection $\{f_\alpha\}_{\alpha \in A}$ so that $\sup_{\alpha \in A} f_{\alpha}(x)$ is finite and non-measurable

Background

Give an example of a collection of measurable non-negative functions $$\{f_\alpha\}_{\alpha \in A}$$ such that if $$g$$ is defined by $$g(x)=\sup_{\alpha \in A} f_{\alpha}(x)$$, then $$g$$ is finite for all values of $$x$$ but $$g$$ is non-measurable. ($$A$$ is allowed to be uncountable.

Attempt

Let $$A$$ be the Vitali set. Then $$A$$ is not Lebesgue measurable. For each $$\alpha \in A$$, let $$f_{\alpha}(x)=\begin{cases} 1 &\mbox{if } x=\alpha \\ 0 &\mbox{if } x \neq \alpha \end{cases}.$$ Then for each $$\beta \in \mathbb{R}$$, $$\{x:f_{\alpha}(x)>\beta\}\in \{\varnothing, \{\alpha\},\mathbb{R}\},$$ so $$f_\alpha$$ is measurable with respect to the Lebesgue $$\sigma-$$ algebra. However, $$g(x)=\begin{cases} 1 &\mbox{if } x\in A \\ 0 &\mbox{if } x \not\in A \end{cases},$$ which is finite and non-measurable since $$\{x:g(x)>0\}=A.$$

Question

Is my example correct? Specifically, is more work required to show that $$g$$ turns out to be as I have claimed?

• The statement immediately following the definition of $f_\alpha$ isn’t quite right. What you mean is that for each $\beta\in\Bbb R$, $$\{x\in\Bbb R:f_\alpha(x)>\beta\}\in\big\{\varnothing,\Bbb R,\{\alpha\}\big\}\;.$$ Other than that, it looks good. Jun 13, 2015 at 19:04
• your functions $f_\alpha$ and $g$ are correct and solve the problem. however, i do not know, what you mean with $\{x:f_{\alpha}(x)>\beta\}=\{\varnothing, \mathbb{R}, \alpha\}$. Jun 13, 2015 at 19:04
• @supinf I just made an edit...is it now correct? Jun 13, 2015 at 19:09
• Not quite: the possibilities are $\varnothing$, $\Bbb R$, and $\{\alpha\}$. Note that $\{x:f_\alpha(x)>\beta\}$ is a set of real numbers, but $\{\varnothing\}$ and $\{\Bbb R\}$ are not, so they can’t be equal to $\{x:f_\alpha(x)>\beta\}$. Jun 13, 2015 at 19:24
• One thing I would like to add that is small, but you say " $\mathbb{R}$ equipped with Lebesgue measure". There are a few things wrong with this statement. First, is this the domain or range? (I'm assuming both, so it's not a big deal). More importantly however, is that measurability has nothing to do with what MEASURE is on the space, it has to do with what $\sigma$ algebra is on the space. I'm assuming you mean Borel $\sigma$ algebra?
– user223391
Jun 13, 2015 at 19:32

Your proof is correct except specifying in more detail about $$\{x:f_{\alpha}(x)>\beta\}\in \{\varnothing, \{\alpha\},\mathbb{R}\}$$ like $$\{x:f_{\alpha}(x)>\beta\}=\begin{cases} \varnothing & \text{ if } \beta\geqslant1 \\ \{\alpha\} & \text{ if } 0\leqslant\beta<1 \\ \mathbb{R} & \text{ if } \beta<0 \end{cases}$$