If $(f-\tilde f)g=0$ for all $g\in L^q$, then $f=\tilde f$

Question

Let $(\Omega,\mathcal A,\mu)$ be a measure space, $p,q\ge1$ with $p^{-1}+q^{-1}=1$ and $f:\Omega\to\mathbb R$ be $\mathcal A$-measurable with $$\int|fg|\:{\rm d}\mu<\infty\;\;\;\text{for all }g\in L^q(\mu)\tag1.$$ By $(1)$, $$L^q(\mu)\ni g\mapsto fg\tag2$$ is a bounded linear fuctional and hence there is a unique $\tilde f\in L^p(\mu)$ with $$(f-\tilde f)g=0\;\;\;\text{for all }g\in L^q(\mu)\tag3.$$

Can we conclude that $f=\tilde f$?

EDIT: As we can see from this answer, we need to impose further assumptions; but which do we really need?

Answer 1

Assume that we don't have $f = \tilde{f}$ almost everywhere. Then there is an $\varepsilon > 0$ and $E$ with finite measure at least $\varepsilon$ such that $|f-\tilde{f}| \geq \varepsilon$ on $E$. Then $g = \operatorname{sign}(f- \tilde{f}) \chi_E$ is in $L^q(\mu)$ and $$\int (f-\tilde{f}) g d\mu \geq \varepsilon^2 > 0.$$

Edit: Note that this answer assumes that the measure space is semifinite to assume pathological examples. At the point it was written, it was clear from discussion with OP that this is what OP wants.

Answer 2

Actually the answer is yes if you assume something about the measure, for example $\sigma$-finiteness is enough. But it's no to the question as stated, even for $p=q=2$.

Say $X=\{0,1\}$, $\mu(\{0\})=\infty$, $\mu(\{1\})=1$. Let $f=1$. Then $fg\in L^1$ for every $g\in L^2$ but $f\not\in L^2$.

Edit: The question has changed; now the OP acknowledges that some additional assumption is needed and asks what assumption. It's easy to see, as mentioned above, that $\sigma$-finiteness is enough. In fact it's sufficient too assume $\mu$ is semi-finite, and that's exactly right: If there exists $E$ such that $\mu(E)>0$ and $\mu(F)$ is $0$ or $\infty$ for every $F\subset E$ then $f=\chi_E$ gives a counterexample.