Complex Integral Function I'm currently stuck on the following problem: 

Let $\phi:[a,b]\times[c,d]\to\mathbb{C}$ be a continuous function and define $g:[c,d]\to\mathbb{C}$ by $$g(t)=\int_{a}^{b}\phi(s,t)\:ds.$$ Show that $g(t)$ is continuous. 

I know that to show that $g(t)$ is continuous at an arbitrary $\alpha\in[c,d]$, we must show that to each $\epsilon>0$ there corresponds a $\delta>0$ such that $$|g(t)-g(\alpha)|=\bigg|\int_{a}^{b}\phi(s,t)\:ds-\int_{a}^{b}\phi(s,\alpha)\:ds\bigg|<\epsilon$$ whenever $|t-\alpha|<\delta$. I tried to work through this, but I kept getting a $\delta$ which depended on $t$. I'm not sure where to go. Clearly, $\phi$ is uniformly continuous since it is a continuous function on a compact set. I think I need to use this uniform continuity, but I'm not sure how to incorporate this fact into the proof. Thanks in advance for any help!
 A: Let's see.. I'm not very good at this sort of thing so tell me if I'm right: uniform continuity means that the patron on the barstool on our right in the denim jacket with the tattoo of Emmy Noether assures us that for any $\epsilon_1>0$ he can, for the right price, provide us with a $\delta_1=\delta_1\left(\epsilon_1\right)>0$ such that $\left|\phi(s,t)-\phi(s,\alpha)\right|<\epsilon_1$ whenever $\left|t-\alpha\right|<\delta_1$.  
Thus, when the gentleman in the black leather jacket on the barstool to our left smashes a bottle on the bar and screams at us that there is no way that for his $\epsilon>0$ that it's possible that $\left|\int_a^b\phi(s,t)ds-\int_a^b\phi(s,\alpha)ds\right|<\epsilon$ we offer the patron on our right $\epsilon_1=\frac{\epsilon}{b-a}$ along with a packet containing a mysterious white powder and he comes back with $\delta_1\left(\frac{\epsilon}{b-a}\right)$ and we can tell the gentleman on our left with the tattoo of Kurt Friedrich Gödel $\delta=\delta_1\left(\frac{\epsilon}{b-a}\right)$ so that
$$\left|\int_a^b\phi(s,t)ds-\int_a^b\phi(s,\alpha)ds\right|\le\int_a^b\left|\phi(s,t)-\phi(s,\alpha)\right|ds\lt\int_a^b\left(\frac{\epsilon}{b-a}\right)ds=\epsilon$$
