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.


Let $A$ be the Vitali set. Then $A$ is not Lebegue 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 $\{x:g(x)>0\}=A.$


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

  • 2
    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jun 13 '15 at 19:04
  • 2
    $\begingroup$ your functions $f_\alpha$ and $g$ are correct and solve the problem. however, i do not know, what you mean with $\{x:f_{\alpha}(x)>\beta\}=\{\varnothing, \mathbb{R}, \alpha\}$. $\endgroup$ – supinf Jun 13 '15 at 19:04
  • $\begingroup$ @supinf I just made an edit...is it now correct? $\endgroup$ – illysial Jun 13 '15 at 19:09
  • 3
    $\begingroup$ 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 $\{x:f_\alpha(x)>\beta\}$. $\endgroup$ – Brian M. Scott Jun 13 '15 at 19:24
  • 1
    $\begingroup$ 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? $\endgroup$ – user223391 Jun 13 '15 at 19:32

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.