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Non Riemann Integrable multiplication of functions

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Do the following:

    (a) find an interval I and functions f, g: I → R such that f and g are both Riemann integrable, but f g is not Riemann integrable.

    (b) find an interval I and functions f, g: I → R such that f and g are both Riemann integrable, but g ◦ f is not Riemann integrable.


    2. Relevant equations

    Apparently there are a couple ways to define the Riemann integral, so this is the definition we're using. It obviously shouldn't matter for this question, but just for notation I guess.

    For a function f on a compact time domain [a,b]:

    If {Pj} is a sequence of partitions of [a,b] with j intervals Ij.

    We say that f is Riemann integrable if lim as j→∞ of (A+(f,Pj) - A-(f,Pj)) = 0, i.e. [itex]\forall[/itex]ε > 0, [itex]\exists[/itex] N such that [itex]\forall[/itex]j ≥ N |A+(f,Pj) - A-(f,Pj)| < ε.

    Where A+(f,Pj) = Ʃ(sup{f(t)|t[itex]\in[/itex]Ij} length(Ij)).

    And A- is the same but with inf instead of sup.

    For an unbounded function f on a non-compact time-domain T:

    If f is Riemann integrable on any compact subset of T, then we basically just find the limit as the bounds of the integral approach the end of the interval T. If this integral exits, then f is Riemann intagrable.

    3. The attempt at a solution

    I think I managed to do (b), so here it is:

    On I = [0,1)

    g(t) = [itex]\frac{1}{t + 1}[/itex] and f(t) = t - 1

    Clearly these are both Riemann integrable on [0,1).

    But then (g ◦ f)(t) = [itex]\frac{1}{t}[/itex]

    So ∫1/t dt = [ln|t|]from a to 1 = -ln(a) for 0<a<1.

    Then lim as a→0 of -ln(a) is not defined.

    Therefore g ◦ f is not Riemann integrable.

    (I don't know if this is sufficient? Does this actually mean g ◦ f is not Riemann intagrable or is it just unbounded? Or are those equivalent? We just started this subject so I don't understand it that well yet.)

    For (a):

    I basically have no idea, I know it can't be done on a compact set, but apart from that I have no idea.

    Any help for this would be greatly appreciated!
     
  2. jcsd
  3. Oct 19, 2012 #2

    jbunniii

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    Your solution for (b) is fine. [itex]g \circ f[/itex] is not Riemann integrable because its graph near x = 0 is too "thick" to have finite area. Putting it another way, the function's value grows too rapidly as x approaches 0.

    For (a), there's a simple example with f = g. Try to choose a function f such f^2 grows too rapidly as x approaches 0, but f grows slowly enough that you can integrate it.
     
  4. Oct 19, 2012 #3

    Dick

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    You've got the first part. Your function is both unbounded and not integrable since the limit doesn't exist. There are functions that are unbounded that ARE integrable. Try t^(-1/3) on (0,1]. Does that give you any ideas for (a)?
     
  5. Oct 20, 2012 #4
    Oh, I didn't think of that at all.

    So if I chose f(t) = t-1/3 and g(t) = t-2/3 or just f(t) = g(t) = t-1/2

    Then (fg)(t) = [itex]\frac{1}{t}[/itex] which is not Riemann Integrable on I = (0,1] right?

    Thanks so much for your guys!
     
  6. Oct 20, 2012 #5

    Dick

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    Sure, those are fine examples.
     
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