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

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.

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I think using Weirstrass approx., or just use the case g=1 on S – gary Aug 24 '11 at 22:38

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 $h(x) g(y)$?
@Nate This is the idea that comes to mind when considering your hint but it seems tedious so let me know if there's a better way. Approximate indicator functions by functions in $C(S)$ which are in turn approximated by functions of the form $h(x)g(y)$ as you say. This is done in the $L^{1}$ norm. This leads to $\int_{S}\mathbb{1}_{B}f=0$ for all Borel sets $B$. This implies $\int_{S}|f|=0$ after considering that $\int_{S}\mathbb{1}_{f\geq 0}f=0$ and $-\int_{S}\mathbb{1}_{f<0}f=0$. Now it follows that $f=0$ a.e. on $S$. – Bones Aug 25 '11 at 1:31
@Bones: First, note that it's linear combinations of functions of the form $h(x)g(y)$ that are dense in $C(S)$. Also, $L^1$ approximation is not quite what you want, but a.e. approximation would work; you can take the approximating sequence to be uniformly bounded (why?) and then use dominated convergence. Otherwise, what you say is pretty much equivalent to what I had in mind. – Nate Eldredge Aug 26 '11 at 2:35