If $f, g \in L_{loc}^1(\Omega)$ and $\langle f, \phi \rangle=\langle g, \phi \rangle$ for all $\phi \in D(\Omega)$, then $f=g$ a.e. in $\Omega$

Question

Let $f, g \in L_{loc}^1(\Omega)$ be locally integrable function. Denote $\langle f, \phi\rangle$ for the action of $f$ on the test function $\phi$ as a distribution. Suppose $\langle f, \phi \rangle=\langle g, \phi \rangle$ for all test functions $\phi \in D(\Omega)$, then $f=g$ a.e. on $\Omega$.
[proof] Let $\Omega_\epsilon = \Omega \setminus \cup_{x \in \partial \Omega} \overline{B_\epsilon(x)}$, where $\partial \Omega$ is the boundary of $\Omega$. Let $\phi_x(y)=j_\epsilon(x-y)$ for $x \in \Omega_\epsilon, y \in \Omega$, where $j$ is the standard mollifier. Then $$\langle f , \phi_x \rangle= \int_\Omega f(y)\phi_x(y)dy=\int_\Omega f(y)j_\epsilon (x-y)dy=\int_{\mathbb{R}^n}f(y)j_{\epsilon}(x-y) \rightarrow f \text{ in } L^1_{loc} (\text{so a.e. on } \Omega)$$ Similarly, $\langle g, \phi_x \rangle \rightarrow g$ in $L^1_{loc}$ (so a.e. on $\Omega$). Hence $f=g$ a.e. on $\Omega$.

I don't get few things in the poof.

1. $\int_\Omega f(y)j_\epsilon (x-y)dy=\int_{\mathbb{R}^n}f(y)j_{\epsilon}(x-y)$
   Here we are extending $f(y)=0$ if $y \in \mathbb{R}\setminus\Omega$, but why does choosing $x \in \Omega_\epsilon$ guarantee that the support of $j_\epsilon(x-y)$ is contained in $\Omega$?

2. We have $\langle f, \phi_x \rangle=f\ast j_\epsilon \rightarrow f$ in $L^1_{loc}$ as $\epsilon \rightarrow 0$ from approximation by $C^\infty$-function. (I think theorem gives us the convergence in $L_1$, but anyway this still holds), but how can we get the pointwise a.e. convergence? I think the convergence in $L_p$ in general only guarantee the pointwise a.e. convergence of the subsequence. Our condition is even weaker. (the convergence in $L^1_{loc}$)

I appreciate your help in advance.

Answer 1

1. $j_\epsilon(x-\cdot)$ is radially symmetric with center $x$. The distance from any $x\in\Omega_\epsilon$ to points in the complement of $\Omega$ is $\ge\epsilon$ therefore $j_\epsilon(x-y)=0$ if $y\notin\Omega$.

2. Choose the subsequence along which $\langle f, \phi_x\rangle\to f$ pointwise a.e. From that subsequence, extract a subsequence along which $\langle g, \phi_x\rangle\to g$. Since the limit of a subsequence is always equal to the limit of the mother sequence, $f=g$ a.e.