2 added 503 characters in body

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

$$|y-y_0|<\delta \Rightarrow \sup_{x\in K}\left|\partial^\alpha_x\bigl(\; \phi_y(x)-\phi_{y_0}(x)\;\bigr )\right| <\ve.$$

1

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

$$\sup_{x\in K, \phi\in \Phi} \partial^\alpha_x\phi(x) <\infty.$$