Riemann Integrability and Jordan Measure It can be proven that if a function, $f : [a,b] \rightarrow \mathbb{R}$ is Riemann-integrable then its graph is measurable and has Jordan measure $0$. Is the converse, for a function true?
That is; 
If for a function $f : [a,b] \rightarrow \mathbb{R}$, the set $\{(x,f(x)) | x \in [a,b]\}$ is Jordan measurable with measure $0$, does it follow that $f$ is Riemann-integrable.  

HINT: 
The graph of  a function with a finite image has  Jordan measure $0$. 
$\bf{Added:}$ After we have seen that the answer to the OP question is negative ( since every function with say an image $\subset \mathbb{R}$ of Jordan measure $0$ has also the graph of Jordan measure $0$), there is still a  result to be answered in the affirmative. Assume for simplicity that $f$ is positive. Then $f$ is Riemann intégrable if and only if the subgraph $\{ y \le f(x) \}$ is Jordan measurable, and the integral of $f$ is the measure of the subgraph. The proof is not difficult. 
