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Math Help - Integrable Functions

  1. #1
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    Integrable Functions

    I need to find 2 functions on [0,1]: one that is integrable, one that isn't, such that their product is integrable.

    I was thinking of using
    g(x) := x-1
    h(x) := 1/(x-1)

    ... But I'm having trouble proving h(x) is not integrable on [0,1]...

    I used the fact that g was monotone to prove it is integrable and obviously gh is integrable because it's constant. Right now we're looking at Riemann integrals, but I don't know how to state the proof...

    Please help!
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  2. #2
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    Let f(x) = 1 if x is rational
    = 0 if not

    let g(x) = 0 fr all x


    then f is not integrable, g is and there product is the constant function 0 which is cleArly integrable
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Let f(x) = 1 if x is rational
    = 0 if not

    let g(x) = 0 fr all x


    then f is not integrable, g is and there product is the constant function 0 which is cleArly integrable
    Almost exactly what I was going to put! Consider f(x)=\begin{cases} -1 & \mbox{if} \quad x\in\mathbb{Q} \\ 1 & \mbox{if}\quad x\notin\mathbb{Q}\end{cases}. Clearly we have that this is not integrable since \sup_{E\subseteq[0,1]}f(x)=1,\inf_{E\subseteq[0,1]}f(x)=-1 for any E. But, f^2=1.

    P.S. The user probably meant non-constant. Otherwise just multiply anything by 0
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MatthewD View Post
    h(x) := 1/(x-1)

    ... But I'm having trouble proving h(x) is not integrable on [0,1]...
    \lim_{x\to1^{-}}h(x)=-\infty what is the first assumption when attempting to integrate a function?

    Spoiler:

    That it's bounded!!!
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  5. #5
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    Sorry for not being more descriptive, but the functions are not supposed to be constant--I apologize for neglecting to post that!

    I see what you're saying: since 1/(x-1) is unbounded, it is not integrable on [0,1]. Correct? Is it enough to say that, or do I need to bring in 2 partitions converging to different points?

    Just curious, for the function f(x):=1 if x rat, f(x):=-1 if x irrat, since it is clearly bounded, wouldn't I have to talk about 2 different partitions? It's frustrating because I know it's not integrable, but since we are only working with Riemann integrable functions and their proofs, I'm not sure what I'm allowed to use: right now we're doing everything with partitions... any ideas?

    I really appreciate your help!
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MatthewD View Post
    Sorry for not being more descriptive, but the functions are not supposed to be constant--I apologize for neglecting to post that!

    I see what you're saying: since 1/(x-1) is unbounded, it is not integrable on [0,1]. Correct? Is it enough to say that, or do I need to bring in 2 partitions converging to different points?

    Just curious, for the function f(x):=1 if x rat, f(x):=-1 if x irrat, since it is clearly bounded, wouldn't I have to talk about 2 different partitions? It's frustrating because I know it's not integrable, but since we are only working with Riemann integrable functions and their proofs, I'm not sure what I'm allowed to use: right now we're doing everything with partitions... any ideas?

    I really appreciate your help!
    I told you exactly why it's not integrable. Let P=\left\{x_0,\cdots,x_n\right\} be any partition of [0,1]. Then, M_j=\sup_{x\in[x_{j-1},x_j]}f(x)=1,N_j=\inf_{x\in[x_{j-1},x_j]}f(x)=-1 and so U(P,f)=\sum_{j=0}^{n}M_j\Delta x_j=\sum_{j=0}^{n}\Delta x_j=\left[(x_1-x_0)+(x_2-x_1)+\cdots+(x_n-x_{n-1})\right]=1. Similarly, you will find that L(P,f)=-1. Thus, U(P,f)-L(P,f)=2 for ANY partition. See why it isn't integrable now?
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