Dini condition: There is a function that satisfies the Dini condition so it means: $\phi(t)=\frac{f(x_0+t)-f(x_0)}{t}$ so $\int_\limits{0}^{\alpha}|\phi(t)|dt<\infty$.
Absolutely continuous: $f$ is also absolutely continuous which translates into: $f\in [a,b]$,$\forall\epsilon>0\exists\delta>0$ such that $[\alpha_i,\beta_i]\subset[a,b]\:\:\:\sum_{i=1}^{n}[\alpha_i,\beta_i]<\delta\implies \sum_{i=1}^{n}|f(\beta_i-\alpha_i)|<\epsilon$
Notes: $\int_\limits{0}^{\alpha}\frac{|f(x_0+t)-f(x_0)|}{t} dt\leqslant \int_\limits{0}^{\alpha}\frac{1}{t}\int_\limits{x_0}^{x_0+t}|f'(y)|dydt\leqslant\mu\int_\limits{0}^{\alpha} 1 dt=\mu\delta$.
Questions:
1) As far as I understand the absolute continuity implies that $f'(t)\leqslant M$ where $M$ is a constant. Does $\mu$ incorporate the derivative or is $|f'(y)\leqslant 1|$?
2) Considering $\int_\limits{0}^{\alpha}\frac{|f(x_0+t)-f(x_0)|}{t} dt\leqslant \int_\limits{0}^{\alpha}\frac{1}{t}\int_\limits{x_0}^{x_0+t}|f'(y)|dydt$. What is the reason of this first inequality? It seems to me as if it was instead an equality.
Thanks in advance!