Self-referential $\varepsilon$-$\delta$-proof? I am trying to show that if $f:[a,b]\to\mathbb R$ is continuous then $F(x) = \int_a^x f(t) dt$ is continuous. Here goes:
Let $x_0 \in [a,b]$ and $\varepsilon > 0$. Since $f$ is continuous there is $\delta > 0$ s.t. $|x-x_0|<\delta \implies |f(x) -f(x_0)| < {\varepsilon \over \delta}$. let $x \in B(x_0, \delta)$. Then 
$$ |F(x) - F(x_0) | = |\int_a^x f(t) dt - \int_a^{x_0} f(t) dt| \stackrel{wlog \ x \ge x_0}{=} |\int_{x_0}^x f(t) dt| < \delta \cdot {\varepsilon \over \delta} = \varepsilon$$
Is this proof with "$\delta$ so that $|f(x) -f(x_0)| < {\varepsilon \over \delta}$ valid? Can it depend on the delta? What about the rest of the proof? Thanks for help.
Could I fix the argument (in the comment)?
 A: There is no problem with putting a $\delta$ in the denominator, since without loss of generality we may assume that $\delta\lt 1$.  With this understanding,  the assertion that something is $\lt \frac{\epsilon}{\delta}$ is weaker than the assertion it is less than $\epsilon$.
A: There are several problems with your proof attempt. 
First of all, indeed the dependence on $\delta$ is invalid. Continuity amounts to: For given $\epsilon$, one can find a $\delta$. But you changed the $\epsilon$ in the middle of the process... 
(As André Nicolas rightly points out, for suitable $\delta$ we may ensure $\epsilon < \frac\epsilon\delta$. In general it is however a bad idea to have a dependence on $\delta$ in your $\epsilon$.)
Furthermore, your final integral estimate is not quite correct. A more suitable attempt would be:
$$\left| \int_{x_0}^x f(t) \,\mathrm dt\right| \le \int_{x_0}^x |f(t)|\,\mathrm dt \le (x-x_0)\sup_{y \in [x_0,x]} |f(y)|$$
I will leave it to you to bound this last quantity; one last hint is that given an $\epsilon_F$ for $F$, we will need to apply continuity of $f$ to a different, suitably chosen $\epsilon_f$.
