Suppose $L$ has a regular parametrix . Assume $U$ is a distribution given in an open set $\Omega \subset R^d$ and $L(U)=f$ , then $U$ is $C^{\infty}$ 
Suppose $L$ has a regular parametrix . Assume $U$ is a distribution given in an open set $\Omega \subset R^d$ and $L(U)=f$ , with $f$ a $C^{\infty}$ function in $\Omega$ , then $U$ agrees with a $C^{\infty}$ function on $\Omega$ 

I have learned this theorem in Stein's functional analysis Page$_{133}$ and the author proved this theorem in the following step .  
$(1)$ Show that it suffices to prove $U$ agrees with a $C^\infty$ function on any ball $B$ with its closure $B^* \subset \Omega$ .  
$(2)$ Show that for any ball $B$ with its closure $B^* \subset \Omega$ , we can find a function $f_B \in C^{\infty}$ which depends on the ball $B$ such that $U$ agree with $f_B$ on $B$ .  
After finished the above two step , the author state that the theorem is proved , but I did not see it .  
My attempt :
For step $(1)$ , assume $U$ argees with $g$ . We need to prove for all $\varphi \in D(\Omega)$ , we have $$U(\varphi)=\int g(x)\varphi(x) \,dx$$ 
If this is not true for some $\varphi$  . We can find a function $\phi$ such that $L\phi=\varphi$ . Then 
$$\int g(x)\varphi(x) \,dx \neq U(\varphi)=L(U)(\phi)=\int f(x) \phi(x) \,dx$$ 
WLOG , we have $g(x)\varphi(x) \lt f(x) \phi(x)$ in some open ball $O$ . Then $$\int g(x) \varphi(x) \chi_O(x) \,dx \lt \int f(x)\phi(x) \chi_O(x) \,dx$$
Notice that $\varphi(x)\chi_O(x)=L(U)(\phi(x)\chi_O(x))$   . However , these two functions are not in $C^\infty$ . So I want to show $$U(\varphi(x)\chi_O(x))=\int g(x) \varphi(x)\chi_O(x) \,dx$$ 
and $$L(U)(\varphi(x)\chi_O(x))=\int f(x) \varphi(x)\chi_O(x) \,dx$$ 
My question :
$(a)$ Can we show step $(1)$ follow the discussion above ? If we do not have the assumption $L(U)=f$ , does $(1)$ still valid ?  
$(b)$ In part $(2)$ , for each ball $B$ , we have different functions $f_B \in C^{\infty}(\Omega)$ , so how to use $(2)$ to show $U$ agrees with a function $f$ ?
 A: For each ball $B_{\alpha} $ , let $f_{\alpha} $ denote the function which agrees with distribution $U$ on $B_{\alpha}$ . Since $\Omega$ is open , it can be write as a union (might not disjoint) of open balls $\bigcup B_{\alpha}$ . We now define the function $f$ as follows .
For each $x$ , we can find a ball $B_{\alpha}$ contains $x$ ,and we define $f(x)=f_{\alpha}(x)$ . To see $f$ is well defined , WLOG, assume $B_{\beta}$ is another ball contains $x$ and $f_{\beta}(x)\lt f_{\alpha}(x)$ . Then we can find a ball $C$ contains $x$ with $C \subset B_{\alpha}$ and $C\subset B_{\beta}$ such that whenever $t \in C$ we have $f_{\beta}(t)\lt f_{\alpha}(t)$ . Find a positive function $\phi \in D(\Omega)$ with $\phi(x)\gt 0$ and supported in $C$ , then we have $$U(\phi)=\int f_{\alpha}(t)\phi(t) \, dt=\int f_{\beta}(t)\phi(t) \, dt$$ which is impossible . So the function $f$ is well defined . Also , note that for each $x\in \Omega$ , we can find a ball $B_{\alpha}$ contains $x$ , and $f(x)=f_{\alpha}(x)$ whenever $x \in B_{\alpha}$ . So $f \in C^{\infty}(\Omega)$ .  
To see $$F(\varphi)=\int f(x)\varphi(x)\,dx$$ whenever $\varphi \in D(\Omega)$ . Since the support $K$ of $\varphi$ is compact , we can find finite union of balls to cover $K$ , we say $K\subset \bigcup_{n=1}^N B_n$ . Then by the partition of unity , we can find function $\eta_k$ $(1\le k\le N)$ $\in D$ with $supp(\eta_k)\subset B_k$ and $\sum_{n=1}^N\eta_n(x)=1$ whenever $x\in K$ . Then we have $$F(\varphi)=F(\sum_{n=1}^N \varphi(x)\eta_n(x))=\sum_{n=1}^N F( \varphi(x)\eta_n(x))=\sum_{n=1}^N \int \varphi(x) \eta_n(x) f(x) \,dx \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\int f(x)\varphi(x) \,dx$$ And the proof is completed .
