Proof of continuity of parametric integral 
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.
 A: 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.
