Let $f$ be integrable on $[a,b]$ and suppose for each integrable function $g$ defined on $[a,b]$, $\int^{b}_afg=0$, then $f(x)=0,\forall x\in[a,b]$ I do not think this is true,
but at the same time I am not sure.
I know that if we assume that f is continuous instead of integrable then this statement is true. I just do not know how to provide a counterexample if it is false to show that this is wrong. Integrable does not imply continuity I know that much.

Another counter example is that consider $f:[0,1]\rightarrow [0,1]$ given by $f(x)=0$ if $x$ is irrational or $x=0$ and $f(\frac{p}{q})=\frac{1}{q}$ where $p\in \Bbb Z-\{0\},q\in \Bbb N,gcd(p,q)=1$ , then $f$ is Riemann integrable and $\int_0^1 f=0$ and for any other Riemann integrable $g$ we have using Cauchy-Schwarz inequality $$|\int_0^1 fg|^2\leq\int_0^1 f^2 ×\int_0^1 g^2\leq \int _0^1 f×\int_0^1 g^2=0×\int_0^1 g^2=0$$ ,hence $\int_0^1 fg=0$ . But $f$ is not identically zero in $[0,1]$. 
Though I used $[0,1]$ as a special interval , by slide modifications you can give argument for general compact interval.Note one thing is that I have only consider Riemann integration i.e. here is no Lebesgue integration. You can prove using only definition of Riemann integration that $\int_0^1 f=0$
