Prove that $(\mu \times \mu)(G) = 0$ where $\mu$ is the Lebesgue measure using Fubini 
Suppose $f: [0,1] \rightarrow [0,1]$ is a Borel measurable function. Suppose $G = \{(x,y)\mid f(x) = y\} \subset [0,1] \times [0,1]$. Use Fubini to prove that 
  $$
(\mu \times \mu)(G) = 0.
$$


What I've tried:
$$(\mu \times \mu)(G) = \iint_{G}\  d\mu d\mu = \iint 1_G(x,y)\  d\mu d\mu.$$
I kind of get stuck here, how am i supposed to go further, is it for example smart to take the rieman integral, or can i say the following thing:
$$G \subset \{ 0\leq x \leq1, f(x)\leq y \leq f(x) \}$$ so we see that
$$\iint 1_G(x,y) d\mu d\mu \leq \iint 1_{\{ 0\leq x \leq1, f(x)\leq y \leq f(x) \}} = \int_{[0,1]} \int_{\{f(x)\}} 1_G(x,y) d\mu d\mu = 0.$$
Since $\{f(x)\}$ is a singleton...?
Kees

Let $G_{x}=\lbrace y\in[0,1]:(x,y)\in G\rbrace$.  Clearly, $\mu(G_{x})=0$ for all $x$ (even a.e. would be sufficient ). Now use the fact that 
$$(\mu\times\mu)(G)=\int_{[0.1]}\mu(G_{x})d\mu(x)=0.$$
