Let $f(x,y)$ be some bounded with its derivatives continuous function on $\Omega \times \overline{\Omega}$, where $\Omega$ is a domain in $\mathbb{R}^n$. Let $f(\,\,\cdot\,,\,y) \in \mathcal{E}(\Omega)$ for any fixed $y$. Let $L\_{x} \in \mathcal{E}'(\Omega)$. Is it true that


$$\int\limits\_{\overline{\Omega}} L\_{x}f(x,y) \, \mu(dy) = L\_{x}\int\limits\_{\overline{\Omega}} f(x,y) \, \mu(dy)$$


holds for any probability measure $\mu$ in $\overline{\Omega}$? If it is true, how to show it? 


If $f(x,y) \in \mathcal{E}(\Omega \times \Omega')$ where domain $\Omega'$ is such that $\overline{\Omega} \subseteq \Omega'$ then the equality holds by virtue of the tensor product of distributions theorem.



Integration under functional sign
Suppose first that $L\_x\in C^\infty\_0(\Omega)$. Then the equality you ask about is Fubini's theorem. 


Suppose now that $L\_x$ is not necessarily smooth. Choose a sequence $\newcommand{\ve}{\varepsilon}$ $L\_{\ve,x}\in \mathscr{E}'(\Omega)$ that converges to $L\_x$ in the weak sense. Then one needs to prove that


$$ \lim\_{\ve\to 0} L\_{\ve,x}\int\_\Omega f(x,y)d\mu(y)=L\_x\int\_\Omega f(x,y)d\mu(y), \tag{A} $$


$$ \lim\_{\ve\to 0}\int\_\Omega (L\_{\ve,x}-L\_x)f(x,y)d\mu(y)=0. \tag{B} $$


The equality (A) is an immediate consequence of the weak convergence. The equality (B) requires an additional assumption on $f$.


Denote by $K$ a compact set containing the support of $L\_x$ and $L\_{\ve, x}$, $\ve$ sufficiently small. If we assume that for any multi-index $\alpha$ we have 


$$ \sup\_{x\in K, y\in \Omega} \partial^\alpha\_x f(x,y) <\infty, \tag{C} $$


then (B) follows by invoking the uniform boundedness principle for $\mathscr{E}'(\Omega)$ which states that if a sequence $u\_n \in \mathscr{E}'(\Omega)$ converges weakly to $0$, then $ u\_n(\phi)\to 0$ uniformly for $\phi$ in a bounded subset of $\mathscr{E}(\Omega)$. 


I recall that a subset $\Phi\subset \mathscr{E}(\Omega)$ is bounded if for any compact $K\subset \Omega$ and any multi-index $\alpha$ we have


$$ \sup\_{x\in K, \phi\in \Phi} \partial^\alpha\_x\phi(x) <\infty. $$


**Update.** Let me set $\phi\_y:=f(x,y)$. To insure the integrability of $y\mapsto L(\phi\_y)$ for any $L\in\mathscr{E}'(\Omega)$ it suffices to assume that the map $\Omega\ni y\mapsto \phi\_y\in\mathscr{E}(\Omega)$ is continuous, i.e., for any $y\_0\in \Omega$, any $\ve>0$, any compact $K\subset \Omega$ and any multi-index $\alpha$ there exists $\delta>0$ such that


$$|y-y\_0|<\delta \Rightarrow \sup\_{x\in K}\left|\partial^\alpha\_x\bigl(\; \phi\_y(x)-\phi\_{y\_0}(x)\;\bigr
)\right| <\ve. $$


