Theorem 1.25: Suppose $ \phi \in L^1(\mathbb{R^n}) $ and $ \int_{\mathbb{R^n}} \phi =1 $ . Also, let $\phi_{\epsilon}(x)=\frac{\phi\left(\frac{x}{\epsilon}\right)}{\epsilon^n}$.Moreover , suppose that : $ \psi(x)= \operatorname{esssup}_{|y|\geq |x|} |\phi(y)|$ is in $ L^1(\mathbb{R^n}) $ and $ f \in L^p(\mathbb{R^n}) $ for some $ p \in [1,\infty] $ Then $ \lim_{\epsilon \to 0} (f \ast \phi_{\epsilon})(x)=f(x)$ for every $ x \in L_f $ where $L_f$ is the Lebesgue set of $ f $ .
In the course of the proof the author obtains for $ x \in L_f$ and for an arbitrary $\delta > 0 $ an $h>0 $ such that : $$ \frac{1}{r^n} \int_{|t|<r} |f(x-t)-f(x)|\mathrm{d}t < \delta , \quad \forall 0<r \leq h $$
Then the author considers the function $$ g(r)= \int_{S^{n-1}} |f(x-ry)-f(x)| \mathrm{d} \sigma_y $$ where $ \sigma $ is the surface measure on $ S^{n-1} $. Then he considers the function $$G(r)=\int_{0}^{r} g(s)s^{n-1} \mathrm{d}s $$
Now the step that I do not quite get is why the following calculation is fully justified : $$ \begin{split} \int_{ |t|<h} |f(x-t)-f(x)| &\frac{1}{\epsilon^n} \psi\left(\frac{t}{ \epsilon}\right) \mathrm{d}t = \int_{0}^{h} r^{n-1}g(r) \epsilon^{-n} \psi\left(\frac{r}{\epsilon}\right) \mathrm{d}r \\ &= \left. G(r) \epsilon^{-n} \psi\left(\frac{r}{\epsilon}\right) \right|_0^h-\int_0^h G(r)\,\mathrm{d}\left( \epsilon^{-n} \psi\left(\frac{r}{\epsilon}\right)\right) \end{split}$$
While it clear for me why $ G $ and $ \psi $ is differentiable ( $ G $ because of the Lebesgue's differentiation theorem and $ \psi $ because it is decreasing [ in $ r $ ] ) I am not sure why the integration by parts formula here is valid. For example I know that if R and Q are both absolutely continuous then we do have that $ \int_{a}^{b} RQ^{\prime}+QR^{\prime}=R(b)Q(b)-R(a)Q(a) $ but I am not sure that this applies here.
Thank you in advance for your help.