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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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