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$\textbf{The Problem:}$ 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.

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    $\begingroup$ Well, your starting postulate in itself is already 80% of the job done. math.stackexchange.com/questions/3213762/… $\endgroup$ – zwim May 4 '19 at 20:58
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    $\begingroup$ How did you get $\dfrac{\varepsilon}{n}$? Isn't it $n\varepsilon$? $\endgroup$ – John Thompson May 4 '19 at 21:03
  • $\begingroup$ @JohnThompson You're right, thanks for noting that typo. $\endgroup$ – Stackman May 4 '19 at 21:05
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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 $m(E)=\sum^{n}_{j=1}(b_j-a_j)<\delta$.

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