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.
$\int_{0}^{1} f(x,y)g(y)dy=0$ for a.e. $x\in [0,1]$ with $g\in C[0,1]$ implies $f=0$ a.e. on $[0,1]\times[0,1]$
1
$\begingroup$
measure-theory
-
0I think using Weirstrass appro$x$., or jus$t$ use the case g=1 on S – 2011-08-24
1 Answers
3
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 $h(x) g(y)$?
-
0@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. – 2011-08-26