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.
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.