TITLE: Prove that $f(t)=0$ for almost every $t\in [0,1]$
QUESTION [6 upvotes]: I would be glad if someone could help me to solve the following exercise.
Let $f:[0,1]\to\Bbb R$ be a bounded measurable function such that $\int_0^1f(t)e^{nt}dt=0$ for every $n=0,1,2,\dots$ Prove that $f(t)=0$ for almost every $t\in [0,1]$.
I know that if $f$ is nonnegative then for $n=0$ the integral $\int_0^1f(t)dt=0$ implies that $f(t)=0$ for almost all $t\in[0,1]$. Thanks!

REPLY [5 votes]: This requires Stone Weiersatrass Theorem. Linear combinations of the functions $\{e^{nt}: n\geq 0\}$ form  a dense set in $C[0,1]$ by this theorem so we get $\int_0^{1} f(t)g(t)dt=0$ for every continuous function $g$. You can now approximate $\chi_{(a,b)}$ by continuous functions and conclude that $\int_a^{b} f(t)dt=0$ whneveer $a <b$. This gives $f=0$ a.e.