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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.

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# Integration under functional sign

Let $f(x,y)$ be some bounded 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.