Prove that $F^{(n)}(z)=\int_{X}\frac{\partial^n f}{\partial z^n}(x,z)\,\mathrm{d}\mu(x)$ This is taken from Problem 4.13 by Christian Berg's Complex Analysis. See also post, not a duplicate! I will copy the problem to show the whole context:src)

4.13. (Requires basic measure theory). Let $(X,\mathbb{E},\mu)$ be a measurable space and let $G\subseteq\mathbb{C}$ be open. Assume that $f:X\times G\to\mathbb{C}$ satisfies
(ⅰ) $\forall x \in X$ : $f(x,\cdot)\in\mathcal{H}(G)$.
(ⅱ) $\forall z \in G$ : $f(\cdot,z)$ is measurable on $X$.
(ⅲ) There exists a measurable function $g:X\to[0,\infty]$ satisfying $\int g\,\mathrm{d}\mu<\infty$, such taht
$$\left|f(x,z)\right|\leq g(x) \quad\text{for}\quad x\in X, \ z \in G.$$
$\mathbf{1^{\circ}}$ Prove that $\frac{\partial f}{\partial z}(\cdot, z)$ is measurable for each $z \in G$.
$\mathbf{2^{\circ}}$ Assume that $\overline{K(z_0,r)}\subseteq G$. Prove that
$$ \left| \frac{\partial f}{\partial z}(x,z)\right|\leq\frac{4}{r}g(x), \quad z\in K(z_0,r/2), \ x \in X.$$
and that
$$\frac{1}{h} (f(x,z_0+h)-f(x,z_0)) = \int_{0}^{1} \frac{\partial f}{\partial z}(x, z_0+th)\,\mathrm{d}t, \quad 0<|h|<r, \ x \in X. $$
$\mathbf{3^{\circ}}$ Prove that
$$ F(z) = \int_{X} f(x, z) \, \mathrm{d}\mu(x), \quad z \in G, $$
is holomorphic in $G$ and
$$ F'(z) = \int_{X} \frac{\partial f}{\partial x}(x, z) \, \mathrm{d}\mu(x), \quad z \in G.$$


Remark. Notice that (ⅲ) can be replaced by local conditions: For each $a \in G$ there exists a disc $K(a, r)\subseteq G$ and a "majorant" $g$, both depending on $a$ such taht
$$\left| f(x, z) \right| \leq g(x) \quad\text{for}\quad x \in X, \ z \in K(a, r).$$
$\color{red}{\blacksquare[}$ Notice also that in this version the results can be applied to $\frac{\partial f}{\partial z}$, so the final conclusion is that we can differentiate the integral infinitely often by differentiating under the integral sign:
$$ F^{(n)}(z) = \int \frac{\partial^n f}{\partial z^n}(x,z) \, \mathrm{d}\mu(x), \quad z \in G, \ n \in \mathbb{N}. \tag*{$\color{red}{]\blacksquare}$} $$

I have solved this problem. The question is right ind the end, see the red part I have marked. I do not know how it is reached? Seems like one should prove it by induction. If I put
$$h(x,z):=\frac{\partial f}{\partial z}(x,z) \quad\text{for}\quad (x,z)\in X\times G, $$
then $h$ clearly satisfies the two first points, but I do not know how to prove the last point (iii). Thank you for your time!
 A: Let $\mu$ be a complex measure on a measurable space
$(X,\mathscr{B})$ and let $G\subset\mathbb{C}$ be open.  Suppose $f$
is a complex valued function in $X\times G$ such that
$f(x,\cdot)$ is holomorphic in $G$ for each $x\in X$, that
$f(\cdot,z)$ is $\mathscr{B}$--measurable for each $z\in G$, and that
$|f(x,z)|\leq g(x)$ for all $(x,z)\in X\times G$  and some $g\in L_1(\mu)$.
Then
$$
F(z):=\int_X f(x,z)\mu(dx),\qquad z\in G.
$$
is well define since $|f|\leq g$.
To show that $F$ is holomorphic in $G$, we will make use of Cauchy's theorem along with dominated convergence.
The conditions of the statement above imply that for $z\in G$ fixed
$$\partial_z f(x,z)=\lim_{w\rightarrow z}\frac{f(x,w)-f(x,z)}{w-z}$$
is measurable in $x$ (is enough to go along sequences $z_n\xrightarrow{n\rightarrow\infty}z$).
For $z_0\in G$, choose $r>0$ small enough so that the  closed ball $\overline{B(z_0;2r)}\subset G$. By Cauchy's formula, for any $a\in B(z_0;r)$ and  $z\in B(a;r/2)$,
$$
\begin{align}
\frac{f(x,z)-f(x,a)}{z-a}&=\frac{1}{z-a}\frac{1}{2\pi i}\int_\gamma\Big(\frac{f(x,\xi)}{\xi-z}-\frac{f(x,\xi)}{\xi-a}\Big)\,d\xi\\
&=\frac{1}{2\pi i}\int_\gamma\frac{f(x,\xi)}{(\xi-z)(\xi-a)}\,d\xi
\end{align}
$$
where $\gamma$ is the path $\gamma(t)=a+re^{it}$, $0\leq t\leq 2\pi$. Then
$$
\begin{align}
\Big|\frac{f(x,z)-f(x,a)}{z-a}\Big|&=\frac{1}{2\pi}\left|\int^{2\pi}_0\frac{f(x, a+r e^{it})}{(a+re^{it}-z)r e^{it}}ir e^{it}\,dt\right|\leq \frac{2g(x)}{r}\tag{1}\label{cf}
\end{align}
$$
since $|a+r e^{it}-z|\geq \tfrac{r}{2}$ for all $t\in[0,2\pi]$. Being a complex measure, $|\mu|(X)<\infty$ and so, we can apply dominated convergence to obtain first that $x\mapsto \partial_z f(x,a)$ is integrable with respect to $\mu$ (with respect to $|\mu|$ rather, which in terms implies integrability with resect to positive and negative parts of the real and imaginary parts of $\mu$); second,  that
$$
\begin{align}
F'(a)=\lim_{z\rightarrow a}\frac{F(z)-F(a)}{z-a}&=\lim_{z\rightarrow a}\int_X \frac{f(x,z)-f(x,a)}{z-a}\,\mu(dx)\\
&=\int_X \lim_{z\rightarrow a}\frac{f(x,z)-f(x,a)}{z-a}\,\mu(dx)=\int_X\partial_z f(x,a)\,\mu(dx)
\end{align}
$$
for all $a\in B(z_0;r)$; and third, $|\partial_zf(x,a)|\leq\frac{2}{r}g(x)$ for all $(x,a)\in X\times B(z_0;r)$.
This shows that $F$ is holomorphic in $G$. As
$|\partial_xf(x,z)|\leq \frac{2}{r}g(x)$ in $X\times B(z_0;r)\subset X\times G$, we can repeat the argument above $n$ times to obtain $F^{(n)}(z_0)=\int_X\partial^n_zf(x,z_0)\,\mu(dx)$ for all $z_0\in G$.
