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Problem on Riemann Integrable functions

  1. Aug 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Let f:[a,b] -> R
    R being the set of real numbers

    If f^3 is Reimann-integrable, does that imply that f is?

    2. Relevant equations

    If f is Riemann-Integrable, then it has upper/lower step functions, such that the difference between the upper and lower sums is less than any [positive] epsilon.

    3. The attempt at a solution

    I'm having a difficult time figuring out what it looks like for f^3 to be Riemann-integrable or, even f^2, for that matter.
  2. jcsd
  3. Aug 4, 2007 #2


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    No it doesn't, but if f is integrable, it would imply that f^3 is. If we add, subtract or multiply two integrable functions, the resultant function is integrable. If we divide an integrable fuction f(x) by g(x) and 1/(g(x)) is bounded on the interval, then f(x)/(g(x)) is also integrable. However, if a function is obtained by a limiting process of a series of approximating integrable functions, the resultant function may not be integrable.
    Last edited: Aug 4, 2007
  4. Aug 4, 2007 #3
    Thanks, bel.

    Can you provide an example (where f^3 is integrable, but f isn't) to backup show the f^3 integrable does not imply f integrable? It's still not clear to me the way f^2 now is (if f is -1 on irrationals, 1 on rationals, f^2 would be integrable, but f wouldn't be, for example).
  5. Aug 4, 2007 #4
    If [itex]h_2:[a,b]\to\mathbb{R}[/itex] is Riemann-integrable and [itex]h_1:\mathbb{R}\to\mathbb{R}[/itex] is continuous, then is [itex]h_1\circ h_2:[a,b]\to\mathbb{R}[/itex] Riemann-integrable?

    Don't take my word on this, I haven't gone through the proof or seen this anywhere, but to me it looks like that the composite function is always Riemann-integrable. If this is true, then you can write [itex]f=(f^3)^{1/3}[/itex], and since mapping [itex]x\mapsto x^{1/3}[/itex] is continuous, this would imply f being Riemann-integrable.
  6. Aug 4, 2007 #5

    Gib Z

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    Homework Helper

    I have completely no rigor on this subject, but I intuitively think that: Say we were to plot the function on some graph, the Riemann Integral fills up the graph with an infinite number of infinitely thin rectangles. Now if we cube rooted the graph, all + and - signs will still be the same, the graph will still be continuous between the points of integration. The new graph is also Riemann Integrable?
  7. Aug 4, 2007 #6
    Thanks! As we've proved [in my class] that continuous functions are integrable, I think this does hold.
  8. Aug 4, 2007 #7
    It would have been smarter for me to ask that are compositions of two integrable functions always also integrable. To me it looks like they always are. If that is incorrect, somebody probably mentions it.

    Continuity was just misleading. For example the [itex]h_1\circ h_2[/itex] itself is not necessarily continuous.
  9. Aug 4, 2007 #8
    The composition of two integrable functions is not necessarily integrable. The one counterexample that immediately comes to mind is not trivial, as it involves the Dirichlet function.

    The Dirichlet function is defined on [0,1] as [itex]f(\frac{p}{q})=1/q[/itex] if [itex]\frac{p}{q}[/itex] is rational and 0 otherwise (p/q in lowest terms). Define g as g(0)=0 and g(x)=1 for all x in (0,1]. The Dirichlet function is integrable on [0,1]. So is g. However, [itex]g \circ f[/itex] is the indicator function of the rationals, which is not integrable.
    Last edited: Aug 4, 2007
  10. Aug 5, 2007 #9
    Okey, I looked it in a wrong way then. Now it looks, that integrability of both f and g, and continuity of f or g, should be sufficient do imply the composition being integrable. Seems like some proving needs to be done, but that's not my problem :biggrin:
  11. Aug 6, 2007 #10
    It's not very difficult if you know what you're doing. Essentially continuity on compact sets gives you uniform continuity. Then you play around, showing that you can get bounds on the oscillations (upper riemann vs lower riemann sums) .
  12. Aug 6, 2007 #11


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    There is a simpler example along the line ZioX was thinking about. Define f(x)=1 if x is rational and f(x)=-1 if x is not. f is not Riemann-integrable, but f^2 is. Keep this example in mind while figuring why the cube is different.
  13. Aug 6, 2007 #12
    In fact this is an example of different thing. ZioX gave an example of two integrable functions, whose composition was not integrable. This instead is an example of how composition of non-integrable and integrable functions can become integrable.
  14. Oct 30, 2010 #13
    f(x)=xsin(1/x) for (0,1]and f(0)=0 and if g:[-1,1]---to R i s riemann integrable prove that gof is riemann integrable?
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