If iterated integral is zero then function is zero We are in Measure & Integration class and were assigned this problem from a chapter on Product Measure & Fubini Theorem:

Let $f$ be a real-valued function, integrable with regards to 2-dimensional Lebesgue measure on $[0, 1]^2$, and also for all $a, b \in [0, 1]$ it is further given that 
  $$\int_0^a \int_0^b f(x.y)\ dy \ dx = 0.$$
  Show that $f = 0 $ almost everywhere.

To me, the problem "looks" intuitive because if $f \neq 0,$ then in the first step the $\int_0^b f(x, y) \ dy \neq 0$ and in the second step $\int_0^a \int_0^b f(x.y)\ dy \ dx \neq 0$, contradicting the given hypothesis. But I do not know to write it down mathematically.
Please help and thank you.

HINT: Use Fubini's theorem and the fact that the double integral is zero for all $a$ and $b$ to show that the integral of the function is 0 on every rectangle in $[0,1]\times [0,1]$. Then prove that any function that is not almost everywhere 0 must have nonzero integral on some rectangle.
