$f: \mathbb{R}^+ \to \mathbb{R}^+$ integrable and $f'$ integrable but $f$ doesn't tend to $0$

Let $$f:\mathbb{R}^+ \to \mathbb{R}^+$$ be a continuous function which is integrable on $$\mathbb{R}^+$$. Ie. $$\int_{\mathbb{R}^+} f \in \mathbb{R}$$. Moreover suppose that $$f'$$ is absolutely integrable, i.e.

$$\int_{\mathbb{R}^+} |f'|\in \mathbb{R}.$$

Then do we have $$\lim_{x \to \infty} f(x) = 0?$$

This should be false, yet I am unable to find a counterexample. I've tried to create the usual counterexample of a function composed of triangles such that the sum of the area of these triangles is a convergent serie but it doesn't work ($$\mid f' \mid$$ is not integrable).

(Note: all the integrals are taken in Riemann sense not Lebesgue).

$$f'$$ absolutely integrable means $$f'$$ integrable, so $$x\mapsto \int_0^x f'(t){\rm d}t=f(x)-f(0)$$ has a finite limit $$\ell$$ for $$x\to+\infty$$.
If this limit were to be different from $$0$$, it should be easy tu prove that $$f$$ can't be integrable on $$\mathbb R^+$$.
• Is this true? Take the Cantor singular function $f$ and extend it to a function $g$ on $\mathbb R^+$ by reflecting it in the line $y=1/2$ and translating it $1$ unit to the right; in general, if $g$ is defined on $[0,n]$, extend it to $[n,n+1]$ by reflecting $gf_{\big|[n-1,n]}$ the line $y=n+1/2$ and translating by $1$ unit. Then, $g'=0$ – Matematleta Nov 11 '18 at 15:27