$f_n$ AC functions s.t $\sum\int_{0}^{1}|f'_n(x)|dx<\infty$ Prove $\sum f_n(x)=f(x)$ converges everywhere and that $f'(x)=\sum f'_n(x)$

Question

Assume $f_n$ is a sequence of AC functions s.t $\sum\int_{0}^{1}|f'_n(x)|dx<\infty$ $f_n(0)=0$

Prove $\sum f_n(x)=f(x)$ converges everywhere and that $f'(x)=\sum f'_n(x)$

MY proof:

$$|f_n(t)|=|\int_{0}^{t}f'_n(x)dx|\leq\int_{0}^{1}|f'_n(x)|$$ Thus $\sum f_n(x)$ converges absolutely.

Now $$\sum f_n(t)=\sum\int_{0}^{t}f'_n(x)dx=\int_{0}^{t}\sum f'_n(x)dx$$ and so taking the derivative $f'(x)=\sum f'_n(x)$ what allowed me to move the sum inside is DCT with the integrable function $\sum|f'_n(x)|$

IS this correct?