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.
(I write this as an answer because it is a bit longer than a comment.)
The problem is with the statement $\delta_k\ast f\to f$ in $L^1$. Why does this happen? You cannot invoke the dominated convergence theorem here.
For example, in Brezis' Functional analysis book this fact is proved using fundamental lemma of variational calculus. Other proofs use the fact that $C_{cpt}(\mathbb{R}^n)$ is dense in $L^1(\mathbb{R}^n)$. This follows from Urysohn's theorem on the abundance of continuous function. This density factn cannot be proved using convolutions, and it is also responsible for the fundamental lemma. Like Nate Eldredge, I believe that the proof of the fundamental lemma you outlined is incorrect.