this is a proof of the fundamental lemma of calculus of variation. Some preparations: Let $g(x):=e^{\frac{-1}{1-||x||}} \chi_{||x||<1},$ with characteristic function $\chi,$ then $$c:= \int_{\mathbb{R}^n}g(||x||^2)dx$$ and $\delta_k(x):=\frac{k^n}{c} g(||kx||^2)$ with $||\delta_k||_{L^1} = 1$ and $supp(\delta_k) = B_{[0,\frac{1}{k}]}(0).$ So we have that $f \in L^1$ has the property that $\delta_k * f \rightarrow f$ in $L^1$. This is basically how you show that $C^{\infty}$ is dense in $L^1.$
The fundamental lemma of the calculus of variation says now that if for $f \in L^1_{loc}(\Omega)$ we have $\int_{\Omega} f\phi dx = 0$ for all testfunctions $\phi$, then we already get $f=0.$
First I assume that $\Omega$ is bounded, because the general case is an easy consequence of this. Now define $\Omega_{\delta}:=\{x \in \Omega: d(x,\partial \Omega) \ge \delta\}$ and $f_m:= f|_{\Omega_m}.$
Now, $f_m * \delta_k \in C_c^{\infty}(\Omega)$ for $k$ large enough and $f_m * \delta_k \rightarrow f_m$ in the $L^1$ sense.
At this point, I know that $0 = \int_{\Omega} f(\delta_k * f_m)$ by the assumption of the fundamental lemma.
I definitely get a subsequence $(\delta_{k_j} * f_m)$ that converges pointwise a.e. to $f_m$ and I want to apply Lebesgue's theorem to conclude
$0 = \int_{\Omega} f(\delta_{k_j} * f_m) \rightarrow \int_{\Omega_m} |f_m|^2$ which shows that $f_m = 0$ a.e. for all m , so $f=0$ a.e., but I don't see why the condition of the dominated convergence theorem is fulfilled here.
Please do not post different proofs to this Lemma, I want to understand this last piece of this one.
The assumptions are not fulfilled in general (ok, they are, since $f=0$ a.e., but we do not want to use that).
Take any $f$ with $f_m \in L^1 \setminus L^2$, for example $f(x)=1/\sqrt{|x|}$ and$\Omega =(-1,1)$. Then $\int |f_m|^2 =\infty$, so that dominated convergence is not applicable.
But you can instead use Fatou's Lemma, which will yield your claim.