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Riemann Integrability Question

  1. Nov 29, 2011 #1
    This isn't a homework question. My adviser has me studying basic analysis and has lately pushed me towards the following question:

    "Let f be any continuous function. Can we prove that there exists a SEQUENCE of step functions that converges UNIFORMLY to f?"

    I have noticed this idea is important to Riemann integrability but I can't seem to prove this. Or notice why the idea of uniform convergence is so key.

    Any help? Thank you
     
  2. jcsd
  3. Nov 29, 2011 #2

    micromass

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    Functions which are the uniform limit of step functions are usually called "regulated functions". The uniform limit is important as it allows us to define a non-ambiguous integral of such a function. We might not do the same for pointswize limits.

    Not all Riemann integrable functions are regulated. Indeed, it is a theorem that the set of discontinuities of a regulated function is countable. It is also known that a function is Riemann integrable iff it's set of discontinuities has Lebesgue measure zero.

    Now, the Cantor set C has Lebesgue measure zero, but is not countable. So the function

    [tex]f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{c} 1~\text{if}~x\in C\\ 0~\text{if}~x\notin C \end{array}\right.[/tex]

    is Riemann integrable but not regulated.

    Your question now asks whether a continuous function [itex]f:[a,b]\rightarrow \mathbb{R}[/itex] is regulated. This is indeed the case.
    A key point here is that f is uniform continuous. So we can determine for each epsilon a delta such that

    [tex]|x-y|<2\delta~\Rightarrow~|f(x)-f(y)|<\epsilon[/tex]

    Now, how do we approximate f by step functions. Well, let [itex][a,b]=[a,a+\delta]\cup [a+\delta,a+2\delta]\cup ...\cup [a+n\delta,b][/itex] be a "partition" such that each part has length [itex]<\delta[/itex]. Then we define

    [tex]g(x)=g(a+k\delta)~\text{if}~x\in [a+k\delta,a+(k+1)\delta][/tex]

    Then the uniform distance between g and f is smaller than epsilon.

    I'll leave the details and verification to you!

    Do not that basically all functions we need in our daily life will be regulated.
     
    Last edited: Nov 29, 2011
  4. Nov 29, 2011 #3

    jgens

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    Micromass beat me to it :( But to add one small point to his post, it is important to note that the compactness of [a,b] is crucial.
     
  5. Nov 30, 2011 #4
    micromass, why is it important that f itself be uniformly continuous? That wasn't one of the assumptions I was given. Shouldn't this work even if f itself is non-uniformly continuous? Thanks for your reply.
     
  6. Nov 30, 2011 #5
    ah I think I answered my own question. It is because it is defined on a compact set, correct?
     
  7. Nov 30, 2011 #6

    micromass

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    Yes: continuous on a compact set is uniform continuous.
     
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