# If $f$ is integrable and bounded over $[\delta,1]$ for all $\delta \in (0,1]$ then is $f$ integrable over all of $[0,1]$ $?$

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.

• do you mean $f:[0,1]\to\mathbb R$ ? – Surb Oct 7 '15 at 17:17
• @Surb : A little elaboration please. Thank you. – user118494 Oct 7 '15 at 17:21
• Surb, please post it as an answer :) – luka5z Oct 7 '15 at 17:21
• 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. – Najib Idrissi Oct 7 '15 at 17:44
• @NajibIdrissi : Either case is possible,I don't know.How does that make dfifferences ? Please elaborate in comment or post in answer . – user118494 Oct 7 '15 at 17:46

## 3 Answers

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.

• Could the downvoter explain ? – Surb Oct 7 '15 at 21:12

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 $[0,1]$.

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.

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