Let $\phi:U\times [a,b]\rightarrow\mathbb{C}$ be a continous function such that for each fixed $t\in[a,b]$ the function $z\mapsto \phi(z,t) $ is holomorphic. Define $F(z)=\int_{a}^{b} \phi(z,t) dt $. I want to prove that $F$ is holomorphic.
My idea is the following: it sufices to show that for each $z\in U$, $F$ is holomorphic in a neighborhood of $z$. We do this by using Morera's theorem. Let $z_0\in U$ and let $B(z_0,r)\subseteq U$. Then for each fixed $t\in[a,b]$ we have that $z\mapsto\phi(z,t)$ is holomorphic in $B(z_0,r)$. Hence, given a triangle $T\subseteq B(z_0,r)$ we get $\int_T \phi(z,t) dz = 0$ for fixed $t\in [a,b]$. I want to use Fubini's theorem so I can "commute" the integrals so that:
$$ \int_T F=\int_T\left[\int_a^b\phi(z,t)dt\right]dz=\int_a^b\left[\int_T\phi(z,t)dz\right]dt=0 $$
Why I think I can use Fubini: $\phi$ is continous on $U\times[a,b]$ and we can assume, WLOG, that $\overline{B(z_0,r)}\subset U$, hence $\phi$ is bounded in $\overline{B(z_0,r)}\times[a,b]$ and thus the integral $\int\int|\phi| |dt| |dz|<\infty$, but I know little of measure theory and I know that there are some subtleties in order to apply the theorem.