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.
 A: Here is a counterexample using Lebesgue integration. 
Let $f=\chi_{\Bbb Q\cap [a,b]}$. Then $f$ is $0$ a.e. , hence Lebesgue integrable. Next for any Lebesgue integrable $g$ we have $fg=0$ a.e. so that $fg$ is Lebesgue integrable and $\int_a^b fg=0$ . But $f$ is not identically zero in $[a,b]$.
A: The conclusion should by $f(x)=0$ almost everywhere. To prove, set $g$ equals the sign of $f$ times $f$, which is integrable and $fg=|f|$. Then you get that $\int|f(x)|dx=0$, which implies that $f=0$ almost everywhere.
A: Okay since this is true for every integrable function g, take $g=f$
Now this becomes $\int^{b}_aff=0$
Now we know that $f^2(x)\geq0$
and $\int^{b}_af^2(x)=0$, This means that the area of the curve is zero even if the function is always above x axis, only one conclusion can be derived from this, that is 
$f^2(x) \equiv 0$ $\forall$ x $\in$ $(a,b)$
The function becomes identically zero.
therefore $f(x)=0,\forall x\in[a,b]$
EDIT: If it is given that f(x) and g(x) is continuous functions then this approach would work, since there is no such condition in this question, hence the statement becomes false.
A: 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 $\int_0^1 f=0$
