Let be $K:M\times [\alpha ,\beta] \to \mathbb{R}$, $(x,t) \mapsto K(x,t)$, where $[\alpha ,\beta]\subseteq \mathbb{R}$ and $M \subseteq\mathbb{R}^n$.
For a fixed $x\in \mathbb{R}^n$ we assume the function $K(x,t):I\to \mathbb{R}$, $t\mapsto K(x,t)$ to be Riemann-integrable. Then, we define the parametric integral by
$F(x):=\int_{\alpha}^{\beta}K(x,t)dt$.
We now have the statement that if $K$ is continuous on $M\times [\alpha ,\beta]$ then $F(x)$, the parametric integral, is continuous on $M$.
When our professor went through the proof of this statement he pointed out that one has to construct a finite covering to make use of the continuity of $K$.
However, I didn't quite understand this point. Why do we need such a finite covering? Maybe someone can explain this to me.
This is the part of the proof where the finite covering comes into play. It is based on the explanation the professor gave me.
When proving the conitnuity of $F(x)$ one has to show that for an arbitrary $\epsilon>0$ there exists a $\delta>0$ for all $t\in [\alpha, \beta]$ such that for all $x \in M$ with $\Vert x -a\Vert<\delta$ it holds $|F(x)-F(a)|\leq \int_{\alpha}^{\beta}|K(x,t) - K(a,t)|dt<\epsilon$.
We now try to show that $|K(x,t) - K(a,t)|<\frac{\epsilon}{\beta-\alpha}$ holds , where $\frac{\epsilon}{2(\beta-\alpha)}>0$. Then, integration would deliver the desired result.
Because of continuity of $K$ we know that for a given $\frac{\epsilon}{2(\beta-\alpha)}>0$ we find a $\delta>0$ such that for all ${x\choose t'}$ with $\Vert{x\choose t'}-{a\choose t}\Vert<\delta \Rightarrow |K(x,t')-K(a,t)|<\frac{\epsilon}{2(\beta-\alpha)}$. We can find such a $\delta(t)>0$ for every $t\in [\alpha, \beta]$. Among these infinitely many $\delta(t)$ we would like to take the smallest one: $\delta_0:=\inf\{\delta(t)~|~t\in[\alpha,\beta]\}$ in order to set $\Vert x-a\Vert <\delta_0$. This would guarantee that there is no $t\in[\alpha,\beta]$ which violates $|K(x,t)-K(a,t)|<\epsilon$. However, it turns out that we cannot exclude that $\delta_0=0$. So, we need a finite covering of $[\alpha,\beta]$ which supplies us with finite many $\delta(t)$ from which we choose the smallest. This $\delta(t)$ is definitely $>0$. This result then helps us to finish the proof.