Statement: If $g$ is any $\Bbb{C}$-valued measurable function on a measure space $(X, \mu)$ such that $$ \int fg d\mu = 0 $$ for all $f\in L^1(\mu)$ then $g=0$ almost everywhere.
Proof. If $\mu$ is $\sigma$-finite then the statement follows from my earlier post; since the condition on $g$ implies that $$ \int_E g d\mu = \int \chi_E g d\mu = 0 $$ for all $\mu(E)<\infty$.
But I'm not sure how to (dis)prove the last "statement" for non-$\sigma$-finite $\mu$. Any help is appreciated. Thanks.
You have missed $\sigma$ finiteness of the measure in formulating your result. There exist measures $\mu$ for which the only integrable functions are $0$ a.e. In this case your new statement is clearly false. [Take $\mu (E)=\infty$ for any non-empty $E$].
If $X$ is an uncountable set, $\mathcal E$ is the $\sigma$-algebra of the countable-or-cocountable subsets of $X$ and $\mu(A)=\begin{cases}0&\text{if }\lvert A\rvert\le \aleph_0\\ \infty&\text{if }\lvert X\setminus A\rvert\le\aleph_0\end{cases}$, then every $L^1$ function is $0$ almost-everywhere, and therefore $\int fg\,d\mu=0$ for all measurable (or even non-measurable) $f$ and for all $g\in L^1$.
