$\int_{0}^{1} f(x,y)g(y)dy=0$ for a.e. $x\in [0,1]$ with $g\in C[0,1]$ implies $f=0$ a.e. on $[0,1]\times[0,1]$

Question

I'm stuck on a measure theory problem and need some hints. Let $S=[0,1]\times[0,1]$ be the unit square in $\mathbb{R}^2$ and $f\in L^1(S)$. Suppose that for any $g$ continuous on $[0,1]$ we have $\int_{0}^{1} f(x,y)g(y)dy=0$ for a.e. $x\in [0,1]$. Then $f=0$ a.e. on S. Thanks in advance.

Answer 1

Hint:

For any $h$ continuous on $[0,1]$, we can say $$\int_0^1 \int_0^1 f(x,y) h(x) g(y) dx dy = 0.$$ What does the Stone-Weierstrass theorem say about the set of all continuous functions on $[0,1]^2$ which are of the form $h(x) g(y)$?