Integration under functional sign 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.
 A: 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
$$|y-y_0|<\delta \Rightarrow \sup_{x\in K}\left|\partial^\alpha_x\bigl(\; \phi_y(x)-\phi_{y_0}(x)\;\bigr
)\right| <\ve. $$
A: Davide's right.  Neither integral makes sense because the function was only continuous. If you assume $f$ is a smooth test function, then a priori it's only clear when $\mu$ is a finite linear combination of point masses.  Thus, you cannot avoid using a Riemann sums trick: approximate the more general measure with a sequence of finitely supported measures $\mu_n$.  My impression is that the general theory of distributions cannot start without taking Riemann-type sums at some point and that any argument probably has this maneuver underlying it somewhere.  
For the right hand side, you need to prove that $\int f(x,y) d\mu_n \to \int f(x,y) d\mu(y)$ in some $C^k$ topology as functions of $x$ over the support of $L_x$.  For the left hand side, check that $\mu_n \rightharpoonup \mu$ weakly and $L_x f(x,y)$ is continuous in $y$.  This step again uses that $L_x$ is continuous with respect to $C^k$ convergence for some $k$.
Note, if you establish this identity when $\mu$ is, say, an absolutely continuous measure with a smooth density function, then you can pass to the limit for a general finite measure by using a mollifying kernel (analogous to taking $L_{\epsilon, x}$ in Liviu's argument, but this is a mollification in the $y$ variable, and is just measure theoretic).
