Differentiating under the (Lebesgue) integral sign. There's this guide for proving the theorem about differentiation under integral sign, I took a look around some questions here, but I still have doubts. I know I can probably find this done in a book, but I'm so close that I'd rather ask here instead.

Let $(Y, {\scr A},\mu)$ be a measure space with $\mu(Y) < +\infty$, $U \subseteq \Bbb R^N$ open, and $f\colon U \times Y \to \Bbb R$ such that $f(x, \cdot)$ is $\scr A$-measurable and bounded, for all $x \in U$, so we can define $F(x) = \int_Y f(x,y)\,{\rm d}\mu(y)$.
(a) If $f(\cdot, y)$ is continuous for all $y \in Y$ and $|f(x,y)| \leq M$ for all $(x,y) \in U \times Y$ for some $M  \geq 0$, then $F$ is continuous.

Let $x_0 \in U$ and take $(x_n)_{n \geq 1} \subseteq U$ such that $x_n \to x_0$. Let's prove that $F(x_n) \to F(x_0)$. By continuity of $f(\cdot, y)$, we have that $x_n \to x_0$ implies that $f(x_n,y) \to f(x_0, y)$. Since $|f(x,y)| \leq M$, we have that $\int_Yf(x_n,y)\,{\rm d}\mu(y) \to \int_Y f(x_0,y)\,{\rm d}\mu(y)$, but that's what we wanted.

(b) Suppose now that $f(\cdot, y)$ is $C^1$ in $U$ for all $y \in Y$ and that $\left|\frac{\partial f}{\partial x_j}(x,y)\right| \leq M_1$ for all $(x,y) \in U \times Y$ for some $M_1 \geq 0$. Show that $F$ is $C^1$ in $U$ and: $$\frac{\partial F}{\partial x_j}(x,y) = \int_Y \frac{\partial f}{\partial x_j}(x,y)\,{\rm d}\mu(y),$$for all $x \in U$, $j=1,2,\ldots, N$.

We can suppose without loss of generality that $N=1$, so we want to prove that $F'(x) = \int_Y \frac{\partial f}{\partial x}(x,y)\,{\rm d}\mu(y)$ here. I also know that once we prove that $F'$ is given by that formula, the proof from (a) applies with $F' \leftrightarrow F$ and $\frac{\partial f}{\partial x} \leftrightarrow f$, so that $F'$ being continuous is a given.
We have to prove the formula, then. All the versions of this exercise I've seen so far or uses some uniform continuity, which I don't have here, or the integral is over $[0,1]$, allowing the ${\rm stuff} = \int_0^1 {\rm stuff}\,{\rm d}y$ step. I only know that $\mu(Y) < +\infty$, not that $\mu(Y) = 1$. The natural thing to do was to use the MVT: \begin{align}\left|\frac{F(x)-F(x_0)}{x-x_0} - \int_Y \frac{\partial f}{\partial x}(x,y)\,{\rm d}\mu(y)\right| &= \left|\frac{f(x,y)-f(x_0,y)}{x-x_0} - \int_Y \frac{\partial f}{\partial x}(x,y)\,{\rm d}\mu(y)\right| \\ &= \left|\frac{\partial f}{\partial x}(\xi, y) - \int_Y \frac{\partial f}{\partial x}(x_0,y)\,{\rm d}\mu(y)\right|,\end{align} for some $\xi$ between $x$ and $x_0$, and I can't do anything from here on. Now what?
 A: The proof is quite similar to the proof of part a). We consider a sequence $x_n \to x_0$ such that $x_n \neq x_0$ for all $n > 0$, and the difference quotient
$$\frac{F(x_n) - F(x_0)}{x_n - x_0} = \int_Y \underbrace{\frac{f(x_n,y) - f(x_0,y)}{x_n - x_0}}_{g_n(y)}\,d\mu(y).$$
By the mean value theorem, for every $n$, and every $y$, there is a $\xi_n(y)$ between $x_n$ and $x_0$ such that
$$\frac{f(x_n,y) - f(x_0,y)}{x_n - x_0} = \frac{\partial f}{\partial x}(\xi_n(y),y),$$
so we have the bound $\lvert g_n(y)\rvert \leqslant M_1$ for all $n$. By the partial differentiability of $f$, we have $g_n(y) \to \frac{\partial f}{\partial x}(x_0,y)$ for all $y$. The dominated convergence theorem thus yields
$$\lim_{n\to\infty} \frac{F(x_n) - F(x_0)}{x_n - x_0} = \int_Y \frac{\partial f}{\partial x}(x_0,y)\,d\mu(y).$$
Since the sequence $(x_n)$ was arbitrary, that means $F$ is differentiable and
$$F'(x_0) = \int_Y \frac{\partial f}{\partial x}(x_0,y)\,d\mu(y).$$
Part a) then yields that $F$ is continuously differentiable.
