We know that if A is dense set in $L^1(\Omega)$. Then if $$\int_{\Omega}fg=0$$ for all $g\in A$ then it is easy to say $f=0$ a.e.
But I am interested to check whether converse also true or false.
A be any subset of $L^1(\Omega )$ with following property:
If $f\in L^{\infty }(\Omega )$ such that $\int fg=0$ for any $g\in A$ then $f=0$.
Then question is that can we prove span of $A$ is dense in $L^1(\Omega)$?
Any Help or hint will be appreciated.
Edit: As s. harp suggested $A$ need not be dense. But span of $A$ is dense .
The closest thing you get to your result as suggested in the comment is the following:
Definition: A subset $A$ in a normed space is called total if the smallest subspace containing $A$ is dense in X. You can take $X = L^{1} (\Omega)$ so that $X^{*} = L^{\infty}(\Omega)$.
Theorem: Prove that $A$ is total if and only if for $f \in X^{*}$ and $\langle f, a \rangle = f(a) = 0$ for each $a \in A$ implies $f =0$.
Proof: Let $M = \overline{ \text{Span} (A)}$. Then $M$ is a closed subspace of a normed linear space $X$. If $A$ is not total, then $M \neq X$. Hence, there exists $x_{0} $ not in $M$. Then by Hahn-Banach Theorem (one of its corollaries), there exists a functional $f \in X^{*}$ such that $f(M) =0$ and $f(x_{0}) = 1 \neq 0$. This proves that if for $f \in X^{*}$ and $f(a) = 0$ for each $a \in A$ implies $f =0$ implies that $A$ must be total.
Now suppose that $A$ is total. Then $X = \overline{ \text{Span} (A)}$. It follows if $x \in X$ then either $x$ is in $A$ or there exists a sequence such that $x_{n} \in A$ such that $x_{n} \to x$ and then by continuity of $f$ together with the fact that $A$ is total imply that if for $f \in X^{*}$ and $\langle f, a \rangle = \int_{\Omega} fa = f(a) = 0$ for each $a \in A$ implies $f =0$.