(1) Let $\Omega$ be an open subset of $\mathbb R^d$ and $f \in L^1(\Omega)$. Does the following assertion hold? What is the measure-theoretic reason if there is one? Counterexamples are welcome.
Assertion: If for any open subset $U\subset \Omega$, we have $\int_U f \ge 0$, then $f \ge 0$ a.e.
(2) The motivation for this issue is due to the question whether it suffices for $f\in L^1(\Omega)$ to satisfy \begin{equation} \int_\Omega fg \ge 0 \end{equation} for all continuous non-negative functions $g$ in order to be a.e. non-negative.
(3) This is true of course if we replace open sets by measurable ones and continuous functions $g$ by measurable bounded ones. If the assertion is true, what is its link to Lusin's theorem? If it is false, how could it be altered to become true (otherwise than stated before)?
Take $\epsilon > 0$ and $n \ge 1$ such that $E := \{x \in \Omega : f(x) \le -\frac{1}{n}\}$ has $|E| > \epsilon$. Take $\delta > 0$ so that for a measurable set $F$ with $|F| < \delta$, it holds that $\int_F |f| < \frac{\epsilon}{2n}$ (this is possible since $f \in L^1$). By basic properties of Lebesgue measure, we can take some open set $U \supseteq E$ with $|U\setminus E| \le \delta$. Then $\int_U f = \int_E f + \int_{U\setminus E} f \le -\frac{1}{n}\epsilon+\frac{\epsilon}{2n} = -\frac{\epsilon}{2n} < 0.$