Proving $f=0$ almost everywhere

This is one of the problem in our past comprehensive exams. I don't mind getting full solution.

Suppose $f$ is a bounded, measurable function on $[0,1]$, $\epsilon>0,$ and for all $x>\epsilon\,$ one has

$$0=\int_0^1 f(s )\exp(-xs)ds$$

Show that $f=0$ almost everywhere.

Someone gave me a hint to solve the problem using Urysohn's lemma. I am not totally comfortable with that lemma. I have a hunch that we can prove this along the line of Fourier analysis. I am not that sure on this approach either. I don't even know how to get started.

• Integral is from $0$ to $1$, I don't know how to incorporate that in the integral sign. I would appreciate if somebody do that for me. – Deepak Dec 20 '12 at 20:46
• is this true for all epsilon or simply a fixed epsilon? If it was a given epsilon, the result may not be true, although I haven't fully thought it through. – toypajme Dec 20 '12 at 21:22
• @toypajme, For me it looks like for all epsilon. This is all the information I have. – Deepak Dec 20 '12 at 21:24

Some hints, hoping they will be useful. Expand the exponential as a power series to deduce that $\int_{[0,1]}f(s)s^nds=0$ for all $n$. This gives, by Stone-Weierstrass theorem that $\int_{[0,1]}f(s)g(s)ds=0$ for all $g$ continuous on $[0,1]$. We conclude from this answer.
• @Deepak Stone-Weierstrass gives that polynomials are dense in the space of continuous functions on $[0,1]$ endowed with the supremum norm. The equality $\int_{[0,1]}f(s)P(s)ds=0$ holds for any polynomial $P$. Using boundedness of $f$ on $[0,1]$, we get for all sequence of polynomials $\{P_n\}$ and all continuous function $g$ that $|\int_{[0,1]}f(s)g(s)ds|\leqslant \sup_{[0,1]}|f|+\lVert g-P_n\rVert_{\infty}$. Now choosing a good sequence we get what we want. – Davide Giraudo Dec 23 '12 at 11:37