If $f\in C^{1}(0,1]\cap C[0,1]$ and $f'\not\in L^{1}(0,1)$, then $f$ oscillates at $0$? Please think it easy because it is not an assignment.

Question : Let $f\in C^{1}(0,1]\cap C[0,1]$ and $f'\not\in L^{1}(0,1)$. Then, does $f$ oscillate frequently and $f'$ is unbounded at $0$?

When I asked the similar question before,  Daniel Fischer taught me that such functions satisfy the above conditions.
I accepted at the time but a abstract proof seems to be difficult after thoughts.
Of course, it is clear that such function is Riemann integrable by the fundamental theorem of calculus and $f'$ should be unbounded at $0$.
The problem is whether $f$ should oscillate at $0$ when thinking in the framework of Lebesgue integral.
I think that an image is what the infinite sum of slope diverges due to oscillation but I don't know well how it can prove.
I'm glad if you give me the strategy of proof when you can prove.
It's good even only hints.
Thank you in advance.

If $f'$ was bounded near $0$, then since $f'$ is continuous on $(0,1]$, then
it would follow that $f'$ is bounded on $[0,1]$ and hence integrable.
Since $\int_x^1 f' = f(1)-f(x)$ for $x>0$, we see that $\lim_{x \downarrow 0} \int_x^1 f'$ exists, hence $f'$ must change sign 'frequently' near zero (if $f'$ does not change sign on $(0,\epsilon)$, then it would be integrable).
The description 'oscillates frequently' is a bit vague. However $f$ cannot
be monotone on any interval $[0,\epsilon)$, otherwise $f'$ would not change
sign (near zero) and would be integrable.
