I found this variant of the fundamental lemma: Let $\Omega \subseteq \mathbb{R}^n$ open and, $f \in L^1(\Omega)$ $$ \int_U f(x)~\mathrm{d}x =0 $$ for every open, smooth set $U \subseteq \Omega$ Then $f =0$ in the $L^1$-sense.
Is my following "proof" correct? Set $$ A_k := \lbrace x \in \Omega: f(x) \geq k^{-1}\rbrace $$ for all $k \in \mathbb{N}$. For every $\varepsilon > 0$ and $k \in \mathbb{N}$ take smooth $U_{\varepsilon, k} \subseteq \Omega$ such that $\lvert U_{\varepsilon, k} -A_k \rvert < \varepsilon$ (this is a standard covering theorem). Then, for fixed $k$: $$ \lim_{\varepsilon \downarrow 0} \int_{U_{\varepsilon, k} -A_k} f(x)~\mathrm{d}x = 0. $$ Moreover $$ \int_{U_{\varepsilon, k} -A_k} f(x)~\mathrm{d}x = \underbrace{\int_{U_{\varepsilon, k}} f(x) ~\mathrm{d}x}_{=0} - \int_{A_k} f(x)~\mathrm{d}x \overset{\varepsilon \downarrow 0}{ \rightarrow} 0 $$ So $$ \int_{A_k} f(x)~\mathrm{d}x = 0 $$ for every $k \in \mathbb{N}$. Passing to the limit $k \rightarrow \infty$ (note that the $A_k$ are increasing, so use montone convergence): $$ \int_{\lbrace x \in \Omega : f(x)> 0 \rbrace} f(x)~\mathrm{d}x = 0 $$ So $\lvert {\lbrace x \in \Omega : f(x)> 0 \rbrace} \rvert = 0$. Pass to $-f$ to prove the result for $f(x)<0$. This concludes the proof.