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?