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?
 A: 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. 
A: Fix $a \in \mathbf R$.
First consider the case that $g$ is bounded in some neighborhood $I$ of $a$. Then there exists $M > 0$ with the property that 
$$ x \in I \implies g(x) \in [g(a) - M,g(a)+M].$$
Call this interval $J$. Since $f$ is continuous and positive on $J$ it attains a positive minimum $m$ on $J$.  Consequently
$$ x \in I \implies |\phi(x) - \phi(a)| = \left| \int_{g(a)}^{g(x)} f(t) \right| \, dt \ge |g(x) - g(a)| m.$$
Let $\epsilon > 0$ be given. Select $\delta > 0$ so that $(a-\delta,a+\delta) \subset I$ and $|\phi(x)-\phi(a)| < m \epsilon$.  Then
$$|x-a| < \delta \implies |g(x) - g(a)| < \epsilon$$
establishing that $g$ is continuous at  $a$.
Second consider the case that $g$ is not bounded on a neighborhood of $a$.  Then there exists a sequence of points $a_n \to a$ with the property that $|g(a) - g(a_n)| \ge 1$ for all $n$.  Without loss of generality we may assume $g(a_n) \ge 1 + g(a)$ for all $n$. Then
$$|\phi(a_n) - \phi(a)| = \int_{g(a)}^{g(a_n)} f \, dt \ge \int_{g(a)}^{g(a)+1} f \, dt > 0$$ for all $n$, so that $\phi(a_n) \not\to \phi(a)$, implying $\phi$ isn't continuous at $a$. Thus $g$ must be bounded near $a$.
A: Suppose that $g$ is not continuous at $y$, there exists $c>0$, a sequence $x_n$ such that $|x_n-y|<{1\over n}$ and $|g(x_n)-g(y)|>c$.
We deduce that $|\phi(x_n)-\phi(y)|=|\int_{g(y)}^{g(x_n)}f(t)dt|=f(u_n)||g(y)-g(x_n)|\geq c|f(u_n)$ where $u_n\in [g(y),g(x_n)]$ or $[g(x_n),g(y)]$.
Since $\phi$ is continuous at $y$, $lim_{n\rightarrow +\infty}|\phi(x_n)-\phi(y)|=0$, we deduce that $lim_{n\rightarrow+\infty}c|f(u_n)|=0$. 


*

*Suppose that the sequence $u_n$ is bounded, there exists a converging subsequence $u_{n_k}$ towards $u$ and $lim_{n_k}f(u_{n_k})=0=f(u)$ contradiction since $f>0$.

*Suppose that $u_n$ is not bounded. This implies that $g(x_n)$ is not bounded, Let $v_n$ a subsequence of $x_n$ such that  $g(v_n)$ is increasing and $lim_ng(v_n)=+\infty,g(v_n)>g(y)$, or $g(v_n)$ is decreasing and $lim_ng(v_n)=-\infty, g(v_n)<g(y)$. Without restriction the generality, we suppose that $lim_ng(v_n)=+\infty$, the other case is similar. We have $|\phi(v_n)-\phi(y)|=\int_{g(y)}^{g(v_n)}f(t)dt>\int_{g(y)}^{g(v_1)}f(t)dt>0$ since $f>0$.
This implies that $lim_nv_n\phi(v_n)\neq \phi(y)$ which is a contradiction with the fact that $\phi$ is continuous and $lim_nv_n=y$.
