When can we switch the limit and the integral? $\Omega$ is a domain in the complex plane and $F(z,t)$ is a continuous function on $\Omega\times I$ where $I=[0,1]$ is the unit interval in $\mathbb{R}$. Suppose further that $F(z,t)$ is analytic in $z$ on $\Omega$ for each fixed $\in I$. Prove that
$$g(z)=\int_0^1 F(z,t) dt$$
Is analytic on $\Omega$. What can be said if $F(z,t)$ is only assume to be analytic on $z\in\Omega$ For all rational values of $t$ (when held fixed in $I$)?
Prove:    $$\lim_{h\rightarrow0} \frac{g(z+h)-g(z)}{h}=\lim_{h\rightarrow 0} \int _0^1 \frac{F(z+h,t)-F(z,t)}{h}$$
Here I want to switch the integral but I do not know which theorem can I use?
Edit 1
Another approach,
Using Morera's theorem,
$\int_{\gamma}\int_0^1F(z,t)dtdz=0$ for any closed curve $\gamma$. Since $F(z,t)$ is uniformly continuous, we can switch the integral and get the result.
Is this right? Can we switch the integral?
Edit 2
Second part: Let $t\in [0,1]$ and let $\{t_n\}$ be a sequence in $\mathbb{Q}$ such that $t_n\rightarrow t$ uniformly. 
$lim_{n\rightarrow\infty}F(z,t_n)=F(z,t)$ uniformly since $F$ is uniformly continuous in $t\in[0,1]$
Then we have: 
$\lim_{n\rightarrow \infty} \int_{\gamma}\int_0^1F(z,t_n)dtdz=\int_{\gamma}\int_0^1lim_{n\rightarrow\infty}F(z,t_n)dtdz=\int_{\gamma}\int_0^1F(z,t)dtdz=0$ by first part.
 A: For the case where $F(z,t)$ is analytic in $z$ for all $t\in I$, if you already knew that $\frac{\partial F}{\partial z}(z,t)$ is continuous - or, if you can use Lebesgue integration theory, that it is (locally with respect to $z$) dominated by an integrable [over $I$] function - you could use a standard theorem about differentiation under the integral sign, e.g. the dominated convergence theorem.
But seeing that one can apply the dominated convergence theorem is not trivial.
One trick to make the legitimacy of the interchange of limit and integration easier to see is to write $F(z,t)$ as a Cauchy integral, so we get
$$g(z) = \int_0^1 \int_{\lvert \zeta-z\rvert = \rho} \frac{F(\zeta,t)}{\zeta-z}\,d\zeta\,dt.$$
Now, if we're looking at $\frac{g(z+h)-g(z)}{h}$ only for $\lvert h\rvert < \rho/2$, since the integrand is a continuous function of $(\zeta,z,t)$, it is easy to see that the convergence of the integrand of the difference quotient $\frac{g(z+h)-g(z)}{h}$ is uniform on the product of the interval and the circle, so interchanging limit and integral is allowed.
This does not work well, if at all, if the analyticity of $F$ in $z$ is only assumed for rational $t\in I$.
Both situations, however, are nicely dealt with by Morera's theorem. By that, $g$, which is known to be continuous, is holomorphic if and only if
$$\int_{\partial\Delta} g(z)\,dz = 0$$
for all triangles $\Delta\subset \Omega$. Inserting the definition of $g$, we find
$$\int_{\partial \Delta} g(z)\,dz = \int_{\partial\Delta} \int_0^1 F(z,t)\,dt\,dz = \int_0^1 \int_{\partial\Delta} F(z,t)\,dz\,dt,$$
where changing the order of integration is allowed since the integrand is continuous and the domains of integration are compact. In the first case, since the inner integral is $0$ for all $t$, Morera's theorem directly asserts the analyticity of $g$.
In the second case, where the analyticity of $F$ in $z$ is only assumed for rational $t\in I$, we note that
$$h(t) = \int_{\partial\Delta} F(z,t)\,dz$$
is a continuous function of $t$, since the integrand is continuous (and the domain of integration compact). Since $h$ vanishes in all rational points, and the rational points are dense in $I$, it follows that $h \equiv 0$ (so in fact, the analyticity of $F$ in $z$ follows for all $t\in I$), and once again, Morera's theorem establishes the analyticity of $g$.
