3
$\begingroup$

Let $f:[0,1]\rightarrow \mathbb R$ be bounded and integrable on $(\delta,1]$ . Is $f$ integrable on $[0,1]$ $?$

I thought of the function ${1}\over {x}$ which could satisfy the condition and prove in negative only it is not bounded on $[0,1]$ , in fact not even defined at $0$ .

I'm unsure where to go from here. Thanks.

$\endgroup$
  • $\begingroup$ do you mean $f:[0,1]\to\mathbb R$ ? $\endgroup$ – Surb Oct 7 '15 at 17:17
  • $\begingroup$ @Surb : A little elaboration please. Thank you. $\endgroup$ – user118494 Oct 7 '15 at 17:21
  • $\begingroup$ Surb, please post it as an answer :) $\endgroup$ – luka5z Oct 7 '15 at 17:21
  • 1
    $\begingroup$ It depends. Is $f$ bounded on $[0,1]$ and integrable on every $(\delta,1]$, or is it (bounded and integrable) on each $(\delta,1]$? The question isn't clear. $\endgroup$ – Najib Idrissi Oct 7 '15 at 17:44
  • $\begingroup$ @NajibIdrissi : Either case is possible,I don't know.How does that make dfifferences ? Please elaborate in comment or post in answer . $\endgroup$ – user118494 Oct 7 '15 at 17:46
4
$\begingroup$

Let $M$ such that $|f(x)|\leq M$ for all $x\in [0,1]$. Let $F(x)=\int_x^1 |f(x)|dx$ with $0<x<1$. The function $F$ is nonincreasing and bounded since $$0\leq F(x)\leq M\int_0^1dx=M,$$ therefore it converges.

$\endgroup$
  • $\begingroup$ Could the downvoter explain ? $\endgroup$ – Surb Oct 7 '15 at 21:12
3
$\begingroup$

The answer is positive: if $f$ is integrable over $[\delta,1]$ for all $\delta \in (0,1]$ then $f$ is integrable over $[0,1]$.

Following is a proof using Riemann integration.

Take $\epsilon > 0$. By hypothesis $f$ is bounded on $[0,1]$, so you can find $M > 0$ with $\vert f(x) \vert \le M$ for $x \in [0,1]$. Take $\delta= \frac{\epsilon}{4M}$. As $\delta > 0$ and $f$ is supposed to be integrable on $[\delta,1]$, you can find two step functions $s,S$ defined on $[\delta,1]$ with $$s(x) \le f(x) \le S(x) \text{ and } \int_\delta^1 (S(x)-s(x)) dx \le \frac{\epsilon}{2}$$

Then extends $s,S$ on $[0,1]$ by defining $s(x)=-M$ and $S(x)=+M$ for $x \in [0,\delta)$. You have $s(x) \le f(x) \le S(x)$ for $x \in [0,1]$, $s,S$ are steps functions and $$\int_0^1 (S(x)-s(x)) dx = \int_0^\delta (S(x)-s(x)) dx + \int_\delta^1 (S(x)-s(x)) dx \le \frac{\epsilon}{2} +\frac{\epsilon}{2}=\epsilon.$$

As this is true for all $\epsilon > 0$, you can conclude that $f$ is Riemmann (hence Lebesgue) integrable on $[0,1]$.

$\endgroup$
1
$\begingroup$

If $f$ is integrable in $[\frac{1}{n},1]$ for every $n$, so the set of descontinuities $C_n$, on $[\frac{1}{n},1]$ has zero measure. Therefore, the set of descontinuities on $[0,1]$ has zero measure and since $f$ is bounded on $[0,1]$ you have that $f$ is integrable.

$\endgroup$
  • $\begingroup$ First sentence is incorrect I think. Or am I wrong? $\endgroup$ – luka5z Oct 7 '15 at 19:01
  • $\begingroup$ $f:[a,b]\rightarrow \mathbb{R}$ is integrable iff the set of its descontinuities has zero measure. $\endgroup$ – Euler88 ... Oct 7 '15 at 19:13
  • $\begingroup$ that is criterion for Riemann integrability... What about $f(x)=\mathbb{1}_{\mathbb{Q}}(x)$ (everywhere discontinuous, yet Lebesgue integrable)? $\endgroup$ – luka5z Oct 7 '15 at 19:23
  • $\begingroup$ My hypotesis is $f$ Riemann integrable. $\endgroup$ – Euler88 ... Oct 7 '15 at 19:56
  • $\begingroup$ I assumed integrable=Lebesgue integrable. OP should specify this. $\endgroup$ – luka5z Oct 7 '15 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.