# Function $f$ from $[0,1]$ to $[0,1]$, bounded, such that the graph of $f$ is not Jordan measurable.

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.

• By the way consider the graph as a subset of R^2 – JSanchez Dec 12 '14 at 4:31
• 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 $[0,1]^2$. – Mirko Dec 12 '14 at 4:39
• Every function into $[0,1]$ is bounded by $2$ – Ross Millikan Dec 12 '14 at 4:42

## 1 Answer

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.

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