Proving the Leibniz integral rule in measure theory The task at hand is to prove that for a function $f\colon X\times (0,1)\rightarrow \mathbb C$ there holds 
$$
\frac d {dt}\int_X f(x,t)\,\mathrm d\mu(x)=\int_X \frac{\partial f}{\partial t}(x,t)\,\mathrm d\mu(x)
$$
at certain conditions, for $X$ a space with measure $\mu$. Namely, the conditions are that:


*

*$f(\cdot, t)\colon X\rightarrow \mathbb C$ is integrable $\forall t\in (0,1)$,

*$\frac{\partial f}{\partial t}(x,t)$ exists for almost all $(x,t)\in X\times (0,1)$,

*$\exists g\in L^1\colon \forall t\in(0,1)\colon\left\lvert\frac{\partial f}{\partial t}(x,t)\right\rvert \leq g(x)$.



Let $F(t)=\int_Xf(x,t)\, d\mu(x)$. Fix $t\in(0,1)$; by the definition of derivative, we have 
\begin{align}
F'(t)&=\lim_{n\rightarrow \infty} n\left(F\left( t+\frac 1 n\right)-F(t)\right)\\
&=\lim_{n\rightarrow\infty}n\left(\int_X f\left(x,t+\frac 1 n\right)\, \mathrm d\mu(x)-\int_X f(x,t)\,\mathrm d\mu(x)\right)\\
&\stackrel{lin.}=\lim_{n\rightarrow \infty}\int_X n\left(f\left(x,t+\frac 1 n\right)-f(x,t)\right)\,\mathrm d\mu(x),
\end{align}
and now let $f_n(x,t):=n\left(f(x,t+\frac 1 n)-f(x,t)\right)$. It's worth noting that $t+\frac 1 n$ may fall out of $(0,1)$, but we can ignore that by defining the expressions above only for large enough $n$ because the limit in $n$ is independent of the first finitely many terms. By assumption, we have that $\frac{\partial f}{\partial t}=\lim_{n\rightarrow \infty} f_n$ exists almost everywhere and therefore so does $\frac{\partial f}{\partial t}(\cdot,t)$. In order to be able to exchange the limit with the integral, we utilize the dominated convergence theorem. However, the condition to use it is $\lvert f_n (\cdot, t)\rvert\leq g_0$ for some integrable function $g_0$ for all $t\in(0,1)$. The assumed function $g$ above is such that $\lvert\lim_n f_n(\cdot,t)\rvert\leq g$, but $(f_n(\cdot,t))$ isn't necessarily increasing in $n$ (for $X=\mathbb R$ we can have a function that is convex in $x$), thus taking $g_0\equiv g$ isn't appropriate.
I've thought of taking $g_0(x)=\sup\{\lvert f_n(x,t)\rvert\;;\;t\in(0,1), n\in \mathbb N\}$, but have no idea how to prove $g_0\in L^1$ then. After finding the function $g_0$ in question, the rest is straightforward.
Additionally, I would like to know if the last condition above can be deduced from continuity of the function $\frac{\partial f}{\partial t}$ in case of $X=\mathbb R$ with Lebesgue measure.
 A: To start, it is natural to me (though unnecessary) to work with continuous limits instead of sequences. You should convince yourself that the dominated convergence theorem applies to this case as well (hint: if the continuous limit is invalid then construct a sequence which violates dominated convergence as you know it). Anyway,
we want to interchange limits in the expression $\lim_{h\rightarrow 0}\int_X\frac{f(x,t+h)-f(x,t)}{h}\,d\mu(x)$. We can use the mean value theorem to relate these difference quotients to the derivative. In particular, we know that for each fixed $x$ such that $f(x,t)$ is differentiable with respect to $t$ that there is some $c\in [t,t+h]$ such that $\frac{f(x,t+h)-f(x,t)}{h}=f'(x,c)$. This would suffice for $f$ real-valued though since $f$ is complex-valued, we may think of it as a vector valued function to $\mathbb{R}^2$ and instead need the mean value inequality to see that $|\frac{f(x,t+h)-f(x,t)}{h}|\leq|f'(x,c)|$. Since we know that $|f'(x,c)|\leq g(x)$, we may apply dominated convergence to conclude.
