Non $\sigma$-finite measure where $\int f g = 0$ for all $g\in \mathrm L^q$ but $f$ is not a.e. zero

Question

On a measured space $(X,\mathcal A,\mu$) with $\mu$ being $\sigma$-finite one can show that if $\int_X fg \ \mathrm d \mu = 0$ for all $g\in \mathrm L^p$, then $f = 0$ $\mu$-a.e.

In an answer here, Danny Pak-Keung Chang provides an example where this is false, but the example is trivial in the sense that here $\mathrm L^p = \{0\}$. Could we come up with a non-trivial example using a non $\sigma$-finite measure ? I've tried for instance using a Hausdorff measure, say $\mathcal H^1$ over $\mathbb R^2$, but could not come up with anything.

Answer 1

Suppose $f$ is a real-valued measurable function such that $\int_X fg\; d\mu = 0$ for all $g \in {\rm L}^p$, but $f$ is not $\mu$-a.e. $0$. There must be $\epsilon > 0$ such that $\mu(f^{-1}((-\infty,-\epsilon) \cup (\epsilon,\infty))) > 0$. WLOG we can assume $\mu(f^{-1}((\epsilon,\infty)) > 0$. Now if $f^{-1}((\epsilon, \infty))$ has a measurable subset $E$ with $0 < \mu(E) < \infty$, we could take $g$ to be the indicator function $I_E$ of $E$ and get $ \int_X f g \; d\mu = \int_E f \; d\mu \ge \epsilon \mu(E) > 0$ while $g \in L^p(\mu)$. So the only possible counterexamples are the "trivial" ones where there is a set of $\mu$-measure $+\infty$ that has no measurable subset with finite and strictly positive measure.