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One part of Theorem 14 (Chapter 5) of the book Real Analysis (3rd edition) by Royden, says that:

if a function $F$ is an indefinite integral, then it is absolutely continuous.

The proof says that this statement is obvious using the following Proposition:

Let $f$ be a non-negative function which is integrable over a set $E$. Then given $\varepsilon \gt 0$, there is a $\delta \gt 0$ such that for every set $A\subset E$ with the $m(A)\lt \delta$, we have $$ \int_A f \lt \varepsilon.$$

I am failing to see how to use the above proposition in the proof and thus will need some help.
Thanks.

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A related question: math.stackexchange.com/questions/40384/… –  Jonas Meyer Jan 26 '12 at 6:30

1 Answer 1

up vote 3 down vote accepted

Assume that $F(x) = F(a) + \int_{a}^x f$ for some integrable function $f \geq 0$.

Let $\varepsilon \gt 0$. By the proposition you mention, there is $\delta \gt 0$ such that $\mu(A) \lt \delta$ implies $\int_A f \lt \varepsilon$. Suppose that $(a_1,b_1), \ldots, (a_n,b_n)$ are non-overlapping intervals of total length at most $\delta$: $$ \sum_{i=1}^n (b_i - a_i) \lt \delta.$$ Then $A = (a_1,b_1) \cup \cdots \cup (a_n,b_n)$ has measure $m(A) \lt \delta$ and $$ \sum_{i=1}^n |F(b_i) - F(a_i)| = \sum_{i=1}^n \int_{a_i}^{b_i} f = \int_A f \lt \varepsilon $$ so that $F$ is absolutely continuous.

The same holds true for arbitrary, not necessarily non-negative integrable functions, you just need to improve the proposition slightly by saying that $\mu(A) \lt \delta$ implies $\int_A|f| \lt \varepsilon$ and the argument above needs a minor adjustment which I leave to you.

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LHF? I almost understand (a. u.) this :) –  The Chaz 2.0 Nov 17 '11 at 1:08
    
heh! :)${}{}{}{}{}{}{}$ –  t.b. Nov 17 '11 at 1:09
    
Thanks very much for your very clear answer. –  Chris Nov 18 '11 at 3:24

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