Let $\varphi: [0,1)\rightarrow [0,\infty)$ be a continuous function such that $\varphi(0)=0$, $\varphi$ is continuously differentiable on $(0,1)$ with $\varphi'>0$ and being nonincreasing on $(0,1)$. My question is:
Whether the function $\varphi'$ is integrable on $(0,1),$ or equivalently, can we write $$\varphi(x)=\int_0^x \varphi'(t)dt \text{ for all } x\in(0,1)?$$
Here the integrand can be understood in the sense of Lebesgue or Riemann. I tried to search some posts about the integrability of the derivative of a continuously differentiable function but still cannot verify the question above.
Edit: Sorry for the mistake it should be $\varphi'$ is nonincreasing on $(0,1)$, not increasing.
Counter-example: $\varphi (x)=-\ln (1-x)$. If you just want integrability on $(0,x)$ for every $x<1$ and your formula to hold, then the answer is YES, as already pointed out in the comments above. Thanks to MW for pointing out that integrability near $1$ is not guaranteed.