Results 1 to 14 of 14

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

  1. #1
    Newbie
    Joined
    Dec 2009
    From
    Brooklyn, NY
    Posts
    6

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

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by evry1cndie View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by evry1cndie View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2009
    From
    Brooklyn, NY
    Posts
    6
    Quote Originally Posted by Drexel28 View Post
    What kind of integral is defined on an open interval?

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

    Quote Originally Posted by Jose27 View Post
    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...



    Quote Originally Posted by mabruka View Post
    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?
    Last edited by evry1cndie; May 12th 2010 at 07:14 AM. Reason: spelling error
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    \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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by evry1cndie View Post


    I'm unfamiliar with this notation, what's lambda in this case? just a change of variable?
    Look here.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by evry1cndie View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by TheEmptySet View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2009
    From
    Brooklyn, NY
    Posts
    6
    .... 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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Dec 2009
    From
    Brooklyn, NY
    Posts
    6
    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by evry1cndie View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Dec 2009
    From
    Brooklyn, NY
    Posts
    6
    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.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by evry1cndie View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Is a bounded integrable function square integrable?
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 28th 2010, 07:26 PM
  2. bounded and continuous function not integrable
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 12th 2010, 12:16 AM
  3. Replies: 3
    Last Post: March 17th 2010, 07:12 PM
  4. Real Analysis: Prove BV function bounded and integrable
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: January 17th 2010, 03:31 PM
  5. Replies: 0
    Last Post: December 1st 2008, 07:43 PM

Search Tags


/mathhelpforum @mathhelpforum