# Graph of a continuous function $f:[a,b]\to(0,\infty)$ is measurable only if $f$ is differentiable?

I am studying measure theory, and I came across the following problem:

Let $f: [a, b]\to (0, \infty)$ be continuous, and let $G = \{ (x, y): y = f(x)\}.$ Prove that $G$ is measurable only if $f$ is differentiable in $(a, b)$.

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## 1 Answer

That claim seems false. In fact, I think it proves itself false, as follows: on the interval $[a,b]$, the function $f:[a,b]\to(0,\infty)$ defined by $f(x)=|x-\frac{a+b}{2}|+1$, which isn't differentiable at $\frac{a+b}{2}$, has a graph of $$G=\{(x,\tfrac{a+b}{2}-x+1)\in\mathbb{R}^2\mid x\in[a,\tfrac{a+b}{2}]\}\cup \{(x,x-\tfrac{a+b}{2}+1)\in\mathbb{R}^2\mid x\in[\tfrac{a+b}{2},b]\}$$ each of which ought to be measurable by the claim, and therefore their union should be too.

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This doesn't contradict the statement since your function takes the value 0 at $x=\frac{a+b}{2}$. – nullUser Jun 24 '12 at 3:45
True, but the problem exists just as well for the function $|x-\frac{a+b}{2}|+1$. I'll edit to reflect your point though. – Zev Chonoles Jun 24 '12 at 4:25
I don't know what the OP's intent is (the graph of a continuous function is always closed hence Borel) but maybe this was sought: It's a non-trivial theorem by Souslin that a function is Borel measurable if and only if its graph is measurable, see e.g. Kechris, classical descriptive set theory, theorem 14.12, page 88 – t.b. Jun 24 '12 at 6:46
I understand the problem given by Zev Chonoles.But G is measurable with respect to which measure? product measure? Also how measurability of G is related to the differentialbility of f? – Madhav Bapat Jun 25 '12 at 16:58