$\textbf{The Problem:}$ Let $f\in L^{1}([0,1])$ and define $$F(x)=\int_{0}^{x}f\quad(x\in[0,1]).$$ Prove directly from the definition of absolute continuity that $F$ is absolutely continuous.
$\textbf{My Thoughts:}$ Since $f\in L^1([0,1])$ we have that for every $\varepsilon>0$ there is a $\delta>0$ such that $$\int_{E}|f|<\varepsilon\quad\text{whenever }E\subset[0,1]\text{ and }m(E)<\delta.$$ Using the above it follows that for all $\varepsilon>0$ there is a $\delta>0$ such that \begin{align*}\sum^{n}_{j=1}|F(b_j)-F(a_j)|&=\sum^{n}_{j=1}\Bigg|\int_{0}^{b_j}f-\int_{0}^{a_j}f\Bigg|\\ &\leq\sum^{n}_{j=1}\int_{a_j}^{b_j}|f|\\ &<\frac{\varepsilon}{n}\\&<\varepsilon \end{align*} whenever $\sum^{n}_{j=1}(b_j-a_j)<\delta$ and the intervals $(a_j,b_j),j,=1\dots,n$ are disjoint. Therefore $F$ is absolutely continuous on $[0,1]$.
Could anyone check if my proof is correct?
Thank you for your time and appreciate any feedback.
Here is a proof: Define $E:=\displaystyle\bigcup_{k=1}^n (a_k,b_k)$.\begin{align*}\sum^{n}_{j=1}|F(b_j)-F(a_j)|&=\sum^{n}_{j=1}\Bigg|\int_{0}^{b_j}f-\int_{0}^{a_j}f\Bigg|\\ &\leq\sum^{n}_{j=1}\int_{a_j}^{b_j}|f|\\ &=\displaystyle\int_E|f|\\&<\varepsilon. \end{align*} since $m(E)=\sum^{n}_{j=1}(b_j-a_j)<\delta$.