function with summable derivative is uniformly continuous?

Question

If we have the function
$f: I \to \mathbb{R}$ with derivative $f’: I \to \mathbb{R}$
such that:
$I \subseteq \mathbb{R}$
$f’ \in L^{1}(I)$
can we deduce that $f$ is uniformly continuous?
Of course it is true if $I $ is compact since $f$ is continuous in $I$ and we can use Heine Cantor theorem; it is also true if $f’$ is bounded. But what happens when $I$ is not compact and $f’$ is unbounded and summable in $I$?
I tried to demonstrate that the implication holds but I failed, so I tried to find a counterexample taking $I$ as a countable union of non-overlapping intervals such that $ I $ is unbounded and $|I| < +\infty $ but I failed again. Can someone help me? Thank you

Answer 1

HINT. It is enough to assume $I=[0, \infty)$ and you can also assume $f(0)=0$. You then have the formula $$\tag{1} f(x)=\int_0^x f'(t)\, dt.$$ Use this to study the increment $f(x)-f(y)$.

IMPORTANT REMARK. As Jose27 pointed out in comments to the main question, it is important to specify the notion of derivative we are talking about. The formula (1) holds if $f'$ is a distributional derivative, provided that $f'$ is an integrable function. It is actually an if and only if. That is, $$ \text{(1) holds} \iff f' \text{ is the distributional derivative of }f \text{ and }f'\text{ is integrable.}$$

In this case, it is said that $f$ is absolutely continuous.

This old question of mine deals exactly with this subject.