If $f$ is integrable in $\mathbb R$, define

$$ F(x)=\int_{-\infty }^xf(t) \, \mathrm dt $$

Is it true that $F$ is Absolutely continuous in $\mathbb R$?

  • $\begingroup$ What have you tried so far? $\endgroup$ – manooooh Mar 22 at 17:07
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    $\begingroup$ Actually I dont have any idea $\endgroup$ – Maryam Amini Mar 22 at 17:09
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    $\begingroup$ And I was looking for some counterexample, but I didnt $\endgroup$ – Maryam Amini Mar 22 at 17:10
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    $\begingroup$ really I didnt find any clue to do that $\endgroup$ – Maryam Amini Apr 12 at 20:06
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    $\begingroup$ Can you accept an answer if some of them was really helpful for you, please? $\endgroup$ – manooooh Apr 16 at 22:24

F is indeed absolutely continuous.

This follows from the more general claim: Let $f \in L^1(\mathbb{R}), \epsilon >0$, then there exists $\delta > 0$ such that any measurable set $A\subset \mathbb{R}$ with $m(A)<\delta$ satisfies $\int_A |f(x)|dx<\epsilon $. Suppose to get a contradiction that there is some $\epsilon > 0$ such that for all $n \in \mathbb{N}$ we have a set $A_n$ such that $m(A_n)<1/n$ and $\int_{A_n} |f(x)|dx \geq \epsilon $. By the dominated convergence theorem we have $\lim_{n \rightarrow \infty} \int_{A_n} |f(x)|dx = 0 $, a contradiction.


It depends on what you mean by "integrable", if you mean Lebesgue integrable then it is true.

If it is Reimann integrable then,

Here is a similar proof sent by @Pedro Tamaroff

One can prove the following

THM Let $f:[a,b]\to\Bbb R$ be Riemann integrable over its domain. Define a new function $F:[a,b]\to\Bbb R$ by $$F(x)=\int_a^x f(t)dt$$ Then $F$ is continuous. That is, the map $$f\mapsto \int_a^x f$$ sends $\mathscr R[a,b]$ to $\mathscr C[a,b]$.

PROOF Let $c\in[a,b]$. Then $$F(x)-F(c)=\int_c^x f(t)dt$$

Since $f$ is integrable, we know it is bounded, say $|f(x)|\leq M$ over $[a,b]$. Then $$ -\int_c^x M\;dt\leq \int_c^x f(t)dt\leq \int_c^xM \;dt$$

which means $$-M(x-c)\leq \int_c^x f(t)dt\leq M(x-c)$$

Thus we get $$|F(x)-F(c)|\leq M|x-c|$$

Taking $x\to c$ the squeeze theorem says $\lim\limits_{x\to c}F(x)=F(c)$. $\blacktriangle$

Another Method:

But also on an interval $[a,b]$

We know that this proposition is true,

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 $$ \int_A f \lt \varepsilon.$$

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

Let $\varepsilon \gt 0$. By the above proposition, 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 less than $\delta$: $$ \sum_{i=1}^n (b_i - a_i) \lt \delta.$$ 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 any not necessarily non-negative integrable functions, we just need to improve the proposition slightly by saying that $\mu(A) \lt \delta$ implies $\int_A|f| \lt \varepsilon$. And you can do nearly the same steps as above to reach the desired result.

  • $\begingroup$ Thank you so much. Is there any other way to do that? $\endgroup$ – Maryam Amini Apr 13 at 14:32

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