Let $k:\mathcal{O}\times\mathbb{R}^n\to\mathbb{R}$, with $\mathcal{O}\subset\mathbb{R}^m$ open, be such that $\forall x\in\mathcal{O}\quad k(x,\cdot)\in L^1(\mathbb{R}^n) $, i.e. the function $y\mapsto k(x,y)$ is Lebesgue summable on $\mathbb{R}^n$, according to the usual $n$-dimensional Lebesgue measure.
I read (theorem 1.d here, p. 2) that
if $\partial_{x_j}\partial_{x_i} k(\cdot,y)\in C(\mathcal{O})$ for almost all $y\in\mathbb{R}^n$ and there exists $ g\in L^1(\mathbb{R}^n)$ such that $$|\partial_{x_j}\partial_{x_i} k(x,y)|\le g(y)\quad\forall x\in\mathcal{O}\quad\text{for almost all }y\in\mathbb{R}^n,$$then$^1$ $$\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$$where $x_i,x_j$ are components of $x$.
I see, by using the corollary to the dominated convergence theorem quoted in this question, that, provided that $\forall x\in\mathcal{O}\quad \partial_{x_i}k(x,\cdot)\in L^1(\mathbb{R}^n) $ (which I do not know how to prove$^2$), then$$\partial_{x_j}\int_{\mathbb{R}^n}\partial_{x_i}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$$but then I do not know how we can "move" $\partial_{x_i}$ outside the integral.
How can we prove that $\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$? I thank any answerer very much.
Notes:
$^1$When I read a text I always assume that it is prefectly worded. Nevertheless, after many fruitless trials, I am not excluding that this result is intended to hold provided that the condition (b) here holds. I would be very grateful to any answerer confirming or denying that.
$^2$By using Fubini's theorem I only see that, once chosen an arbitrary interval $[a,t]$, the function $\partial_{x_i}k(x_t,\cdot)-\partial_{x_i}k(x_a,\cdot)$, where $x_t$ has $t$ as its $j$-th component, is summable and its integral is $\int_{[a,t]}\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y) \,d\mu_y \,d\mu_{x_j}$.
