Let $g:U\times V\subset \Bbb C\times \Bbb C\to \Bbb C$ a continuous function on $U\times V$ with $U\times V$ endowed with the product topology and $\gamma:[0,1]\to V$ a piecewise smooth curve on $U$. Define $\Psi:U\to \Bbb C$ by $$\Psi(z)=\int_{\gamma}g(z,w)dw$$ Prove that $\Psi$ is continuous on $U$.
Try: What I have to prove is that $\displaystyle \lim_{z\to a}\Psi(z)=\Psi(a)$. So, suppose $|z-a|<\delta$. Then, $$\begin{align*}|\Psi(z)-\Psi(a)|&=\left |\int_{\gamma}g(z,w)dw-\int_{\gamma}g(a,w)dw \right |\\ &=\left |\int_{\gamma}(g(z,w)-g(a,w))dw \right|\\ &\le\int_{\gamma}|g(z,w)-g(a,w)||dw|\end{align*}$$
After that I want to use that $g$ is continuous in $U\times V$ in order to bound $|g(z,w)-g(a,w)|$ but I don't know how, any help.
If we further assume that $U$ is open, then it is possible to prove.
$\gamma([0, 1])$ is a continuous image of compact subset $[0, 1]$ of $\mathbb{R}$, which is also compact. Also, the set $A = \{z\in\mathbb{C}: |z-a|\le\delta\}$ is a compact subset of $U$. Hence $A\times\gamma([0, 1])$ is a compact subset of $U\times V$, so that $g$ is uniformly continuous on $A\times\gamma([0, 1])$. Hence by ML inequality, $$|\Psi(z)-\Psi(a)|\le\varepsilon\cdot L(\gamma)$$ where $\varepsilon>0$ is given and we choose $\delta>0$ s.t for $|z-a|<\delta$, $|g(z, w)-g(a, w)|<\varepsilon$ for any $w$(also choose $\delta$ so that $A\subset U$). This completes the proof.