# If $f$ is integrable in $\mathbb R$, Is it true $F$ is Absolutely continuous?

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 $$\mathbb R$$?

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

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