Prove that if $\phi(x) = \int_{0}^{g(x)} f(t) \;dt$ is continuous then $g$ is continuous Considering $g : \mathbb{R} \rightarrow \mathbb{R}$, $f>0$, $f \in C(\mathbb{R})$ and $\phi : \mathbb{R} \rightarrow \mathbb{R}$, As I see we only have to prove $d\!\left(\phi(x),\phi(a)\right)>d\!\left(g(x),g(a)\right) \;\forall_a$.
I’ve got $d\!\left(\phi(x),\phi(a)\right) = 
\left|\int_{0}^{g(x)}f(t)\;dt - \int_{0}^{g(a)}f(t)\;dt\right| = \left|\int_{g(a)}^{g(x)}f(t)\;dt\right| = \left|(g(a)-g(x))f(\zeta)\right|$ 
for some $\zeta$, if $1 \leq f(\zeta) \forall \zeta$ we got it, but I only have $f>0$, any idea about how it follows?, am I on the right way?

You are more or less on the right track. The goal is to bound $|f(\zeta)|$ (locally uniformly) from below, where we need to use the continuity of $f$.
Fix $a\in \mathbb R$, since $f$ is continuous at $a\in \mathbb R$ and $f(a) >0$, there is $\delta_1 >0$ and $m>0$ so that $f(x) > m$ for all $x\in \mathbb R$ with $|x-a|<\delta_1$ (how?). 
With this $m$ and since $\varphi$ is continuous, there is $\delta_2 >0$ so that if $x\in \mathbb R$ and $|x-a|<\delta_2$, then 
$$ |\varphi (x) - \varphi (a) | < m\epsilon.$$
Now choose $\delta = \min\{ \delta_1, \delta_2\}$. Then for all $x\in \mathbb R$ with $|x-a|<\delta$, there is $\zeta$ between $a$ and $x$ so that 
$$ |\varphi (x)-\varphi(a)| = |g(x) - g(a)| |f(\zeta)| \ge m|g(x) - g(a)|.$$
(since we also have $|\zeta - a|<\delta_1$). Thus $|g(x)-g(a)|<\epsilon$ for all $x\in \mathbb R$ with $|x-a|<\delta$. Thus $g$ is continuous. 
