# If $f\in L^{1}([0,1])$ and $F(x)=\int_{0}^{x}f$ for $x\in[0,1],$ prove $F$ is absolutely continuous from the definition

$$\textbf{The Problem:}$$ 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.

• Well, your starting postulate in itself is already 80% of the job done. math.stackexchange.com/questions/3213762/… – zwim May 4 '19 at 20:58
• How did you get $\dfrac{\varepsilon}{n}$? Isn't it $n\varepsilon$? – John Thompson May 4 '19 at 21:03
• @JohnThompson You're right, thanks for noting that typo. – Stackman May 4 '19 at 21:05

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 $$m(E)=\sum^{n}_{j=1}(b_j-a_j)<\delta$$.