2
$\begingroup$

I am currently doing a problem in my real analysis text book and I am having trouble with one in particular. It asks whether or not the graph of a bounded function is Jordan Measurable or not. The function isn't necessarily continuous or continuous almost everywhere continuous it is just bounded. I personally think there does exist a bounded function from $[0,1]$ to $[0,1]$ which has a graph that is not Jordan measurable but I cannot find the function. Any help will be greatly appreciated.

$\endgroup$
  • $\begingroup$ By the way consider the graph as a subset of R^2 $\endgroup$ – JSanchez Dec 12 '14 at 4:31
  • $\begingroup$ From Wikipedia: A bounded set is Jordan measurable if and only if its indicator function is Riemann-integrable. Also, a bounded set is Jordan measurable if and only if its boundary has Lebesgue measure zero. It is not difficult to describe functions form $[0,1]$ to $[0,1]$ such that their graph is dense in $[0,1]^2$ (and hence the boundary of the graph is $[0,1]^2$. $\endgroup$ – Mirko Dec 12 '14 at 4:39
  • $\begingroup$ Every function into $[0,1]$ is bounded by $2$ $\endgroup$ – Ross Millikan Dec 12 '14 at 4:42
1
$\begingroup$

Let $B$ be a Hamel basis of $[0,1]$, i.e. the set of equivalence classes of relation $\sim$ defined by $x\sim y \iff x-y\in\mathbb{Q}$. Then any bijection between (equivalence classes of) $B$ and $[0,1]$ is not Jordan measurable, because it is dense in $[0,1]^2$ and interior of any intersection with any rectangular is empty.

$\endgroup$
  • $\begingroup$ Could you define what exactly a Hamel basis is? $\endgroup$ – JSanchez Dec 12 '14 at 4:44
  • $\begingroup$ @JSanchez Please see my extended answer. $\endgroup$ – Przemysław Scherwentke Dec 12 '14 at 4:48

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.