A proof of Leibniz integral rule

Question

I'm reading Leibniz integral rule

Let $X$ be an open subset of $\mathbb R^d$ and $(\Omega, \mathcal F, \mu)$ a measure space. Let $f:X \times \Omega \to \mathbb R, (x, \omega) \mapsto f(x, \omega)$. We denote by $\nabla$ the gradient w.r.t. the spatial variable $x \in \mathbb R^d$. Assume that

1. $f(x, \cdot)$ is $\mu$-integrable for every $x \in X$,
2. $f (\cdot, \omega)$ is differentiable for $\mu$-a.e. $\omega \in \Omega$,
3. There is a $\mu$-integrable map $\theta:\Omega \to \mathbb R$ such that for every $x \in X$ we have$|\nabla f(x, \cdot)| \le |\theta|$ $\mu$-a.e.

Then for every $x \in X$, $$ \nabla \int_{\Omega} f(x, \omega) \, \mathrm d \mu (\omega) = \int_{\Omega} \nabla f(x, \omega) \, \mathrm d \mu (\omega). $$

Could you have a check (at the end of my proof) if I correctly use the given hypotheses and apply dominated convergence theorem?

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We define $g:X \to \mathbb R$ by $$ g(x) := \int_{\Omega} f(x, \omega) \, \mathrm d \mu (\omega) \quad \forall x \in X. $$

Fix $x \in X$. We need to prove $$ \nabla g (x) = \int_{\Omega} \nabla f(x, \omega) \, \mathrm d \mu (\omega). $$

It suffices to prove that $$ \frac{\partial}{\partial x_1} g (x) = \int_{\Omega} \frac{\partial}{\partial x_1} f(x, \omega) \, \mathrm d \mu (\omega). $$

Let $\{e_1, e_2, \ldots, e_d\}$ be a standard basis of $\mathbb R^d$. Then $$ \begin{align} \frac{\partial}{\partial x_1} g (x) &= \lim_{\varepsilon \to 0} \frac{g(x+\varepsilon e_1)-g(x)}{\varepsilon} \\ &= \lim_{\varepsilon \to 0} \int_{\Omega} \frac{f(x+\varepsilon e_1, \omega) - f(x, \omega)}{\varepsilon} \, \mathrm d \mu (\omega). \end{align} $$

For $\varepsilon \neq 0$, we define $f_\varepsilon: \Omega \to \mathbb R$ by $$ f_\varepsilon (\omega) := \frac{f(x+\varepsilon e_1, \omega) - f(x, \omega)}{\varepsilon} \quad \forall \omega \in \Omega. $$

Then $$ \frac{\partial}{\partial x_1} g (x) = \lim_{\varepsilon \to 0} \int_\Omega f_\varepsilon \, \mathrm d \mu. $$

• By (1.), $f_\varepsilon$ is $\mu$-integrable for all $\varepsilon \neq 0$.
• By (2.), $f_\varepsilon \xrightarrow{\varepsilon \to 0} \frac{\partial}{\partial x_1} f(x, \cdot)$ for $\mu$-a.e.
• By mean value theorem, $|f_\varepsilon| \le \varepsilon \sup_{t\in [0, 1]} |\nabla f(x + t \varepsilon e_1, \cdot )|$.
• By (3.), $|f_\varepsilon| \le \varepsilon |\theta|$ $\mu$-a.e.

By dominated convergence theorem, $f_\varepsilon \xrightarrow{\varepsilon \to 0} \frac{\partial}{\partial x_1} f(x, \cdot)$ in $L^1 (\mu)$. Then $$ \int_\Omega f_\varepsilon \, \mathrm d \mu \xrightarrow{\varepsilon \to 0} \int_\Omega \frac{\partial}{\partial x_1} f(x, \cdot) \, \mathrm d \mu. $$

The claim then follows.