Function Bounded and Continuous on (0,1) but not Integrable!

**evry1cndie** · May 11th 2010, 06:29 AM

Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!

**Drexel28** · May 11th 2010, 09:14 AM

Originally Posted by evry1cndie

Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!

What kind of integral is defined on an open interval?

**Jose27** · May 11th 2010, 12:15 PM

Originally Posted by evry1cndie

Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!

This isn't true for the Riemann integral, since we can extend it to a function in [0,1] as 0 in the endpoints, and it would only be discontinous there.

**mabruka** · May 11th 2010, 12:38 PM

I dont think it is true either, if f is continuous and bounded over (0,1) then

$\bigg|\int_0^1 f(x)dx\bigg|\leq\bigg|\int_{(0,1)} fd\lambda\bigg|\leq M \lambda((0,1))<\infty$

so f would be integrable.

Thus it is important to remove some of the hypothesis:

continuity or boundedness or measure finite set

so we can have some functions satisfying the statement that is left after the removal.

Thats my opinion

**evry1cndie** · May 12th 2010, 06:14 AM

Originally Posted by Drexel28

What kind of integral is defined on an open interval?

I have no idea, unless this was a rhetorical question.

Originally Posted by Jose27

This isn't true for the Riemann integral, since we can extend it to a function in [0,1] as 0 in the endpoints, and it would only be discontinous there.

this was my thought exactly, then if it can be extended to the closed interval it can be integrated and then integrable on all sub-intervals including the open one...

Originally Posted by mabruka

I dont think it is true either, if f is continuous and bounded over (0,1) then

$\bigg|\int_0^1 f(x)dx\bigg|\leq\bigg|\int_{(0,1)} fd\lambda\bigg|\leq M \lambda((0,1))<\infty$

so f would be integrable.

Thus it is important to remove some of the hypothesis:

continuity or boundedness or measure finite set

so we can have some functions satisfying the statement that is left after the removal.

Thats my opinion

I'm unfamiliar with this notation, what's lambda in this case? just a change of variable?

**mabruka** · May 12th 2010, 02:22 PM

$\lambda$ denotes the Lebesgue measure , nevermind then,

the property im using in terms of Riemann integrals :

if f is continuous and bounded then

$\bigg |\int_a^b f(x)dx \bigg|\leq \int_a^b |f(x)|dx\leq \ (b-a)M$

So if f is bounded then you can bound the integral by f's bound multiplying it by the lenght of the integrating interval.

This is a consequence of integral monotonicity, this says:

if $g(x) \leq h(x)$ for all x then

$\int g(x) dx \leq \int h(x) dx$

Using the above for $|f(x)|\leq M$ we get the first one.

**Drexel28** · May 12th 2010, 02:22 PM

Originally Posted by evry1cndie

I'm unfamiliar with this notation, what's lambda in this case? just a change of variable?

Look here.

**TheEmptySet** · May 12th 2010, 07:33 PM

Originally Posted by evry1cndie

Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!

The function

$f(x)=\sin\left( \frac{1}{x}\right)$
is continous on $(0,1)$ and bounded there but is not integrable and cannot be continously extented to 0.

**Drexel28** · May 12th 2010, 07:38 PM

Originally Posted by TheEmptySet

The function

$f(x)=\sin\left( \frac{1}{x}\right)$
is continous on $(0,1)$ and bounded there but is not integrable and cannot be continously extented to 0.

So? Define $f(0)=0,f(1)\sin(1)$ . This is not continuous but surely integrable.

**evry1cndie** · May 12th 2010, 08:59 PM

.... ok, before everyone starts fighting like dogs, the question was improperly phrased. it should have been "prove/disprove."

spoke to my professor today... Thanks all for you're help. I was thinking of the sin(1/x) but also found it was integrable, the only second option was a sawtooth function with base 1/n... but it faced the same fate as sin(1/x) it would surely be integrable... Thanks for you're help guys... sorry if this question made you doubt the integrability of continuous functions. heh.

**evry1cndie** · May 12th 2010, 09:42 PM

Here's my attempt at a proof, any constructive criticism is accepted...

Suppose f is a bounded continuous function on (0,1), then f can be extended to a continuous function g on [0,1].

Since g is bounded and continuous on the closed interval [0,1] it must therefore be integrable.

In the case of f, the worst case scenario is that f is discontinuous at the endpoints, for if it was unbounded at the endpoints it would make a jump and therefore be discontinuous. In which case we have a finite number of discontinuity points (namely two.) But since we are dealing with a finite number of discontinuities the function f is still integrable.

**Drexel28** · May 12th 2010, 09:48 PM

Originally Posted by evry1cndie

Here's my attempt at a proof, any constructive criticism is accepted...

Suppose f is a bounded continuous function on (0,1), then f can be extended to a continuous function g on [0,1].

This is not true at all, TheEmptySet, gave an example above, you would need that $\lim_{x\to0^+}f(x),\lim_{x\to1^-}f(x)$ exist. You would need that $E\subseteq [0,1]$ is closed (viz extension theorem). But, it is true that regardless of what $f$ is you can extend it to $g:[0,1]\to\mathbb{R}:x\mapsto\begin{cases}f(x)\quad\text{ if}\quad x\in(0,1) \\ 0\quad\text{if}\quad x=0,1\end{cases}$ which is bounded, and has finitely many discontinuities and is thus integrable.

**evry1cndie** · May 12th 2010, 09:58 PM

I see, by adding that the points where something may go horribly wrong, like at 0 for sin(1/x), automatically go to 0 we create the discontinuity points and then we're dealing with an integrable function. awesome thanks.

but the remainder of the proof holds true though... right? I just started doubting myself and thinking that I've proved that the extension is integrable, and haven't shown well enough that this implies the original function was integrable.

**Jhevon** · May 15th 2010, 07:34 AM

Originally Posted by evry1cndie

but the remainder of the proof holds true though... right? I just started doubting myself and thinking that I've proved that the extension is integrable, and haven't shown well enough that this implies the original function was integrable.

i agree. here is where i'd probably go back to the definition, maybe the one with the upper and lower sums. maybe let $g$ be the extension such that it is $f$ in the open interval, but $\text{max}|f|$ at the endpoints. then say something like $U_f - L_f \le U_g - L_g < \epsilon$ (for any $\epsilon > 0$ ), since we know $g$ is integrable. This implies f is integrable. but i suck at integration theory, so someone may want to point out a problem with this

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