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Lipschitz Continuity and measure theory

  1. Dec 23, 2006 #1
    Hi, this is not a homework problem because as you can see, all schools are closed for the winter break. But I'm currently working on a problem and I'm not sure how to begin to attack it. Here's the entire problem:

    Let f be bounded and measurable function on [0,00). For x greater than or equal to 0, define F(x)= \int_{0,...,x} f(t)dt.

    Part i. Show that there is some positive M such that |F(x)-F(y)| <= M|x-y|.

    Part ii. Prove that there exist a constant C such that m(F(E)) <= C(m(E)) for every Lebesgue measurable set E of [0,00). Note that m(E) is the Lebesgue measure of the set E and F(E) := {F(x) : x is in E}.


    I know how to do Part i. But as for Part ii., I'm not sure how to begin working on it. Can I rewrite E as an infinite disjoint union of open/closed sets??


    Note that 00 means infinity and <= means less than or equal to.

    Thank you.
     
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  3. Dec 23, 2006 #2

    Hurkyl

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    Yes, but not in a useful way. For example, the Cantor set requires an uncountable union.


    I don't see the proof immediately, but I can see how to do a simpler case. What if f is continuous, and E sufficiently nice? (And, thus, F is differentiable) Remember that [itex]m(S) = \int_S 1 \, dx[/itex].
     
    Last edited: Dec 23, 2006
  4. Dec 23, 2006 #3
    Thanks Hurkyl. I forgot about the Cantor set. I'll think about it some more but I still don't see it.

    We have to find C so that the above inequality works, and I think we're supposed to use Part i.

    Hmm... maybe if we begin by assuming that E is an open set, then |F(x)-F(y)| <= m(F(E))? At this moment I'm not sure how m(E) fits in with m(F(E)).

    Also, whether f is continuous or not, F is absolutely continuous. So F is always differentiable, isn't it?
     
  5. Dec 23, 2006 #4

    Hurkyl

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    F is almost everywhere differentiable, IIRC. (but not necessarily everywhere differentiable)


    The starting point that I'm imagining is that

    [tex]\int_{F(E)} 1 \, dx = \int_E f(x) \, dx.[/tex]

    But that doesn't work in the most general case.

    I think you see something similar; did you mean to say that you're assuming E to be the interval (y, x)? In that case, m(E) = |x - y|, which is how you can get the desired relationship.
     
  6. Dec 23, 2006 #5

    StatusX

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    Start by proving it for an open interval, then an open set, then an arbitrary measurable set.
     
  7. Dec 23, 2006 #6

    AKG

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    If it holds for E an open interval, then it holds for E a countable union of disjoint open intervals by subadditivity:

    [tex]m(F(\cup E_i)) = m(\cup F(E_i)) \leq \sum mF(E_i) \leq \sum Cm(E_i) = C\sum m(E_i) = cm(E)[/tex]

    Thus it holds for any open set. Thus it holds for any countable intersection of nested open sets, since:

    [tex]m(F(\cap U_i)) \leq m(\cap F(U_i)) = \mbox{inf} mF(U_i) \leq \mbox{inf} Cm(U_i) = C\mbox{inf} m(U_i) = Cm(\cap U_i)[/tex]

    Thus it holds for any [itex]G_{\delta}[/itex] set. But every measurable set is just a [itex]G_{\delta}[/itex] set unioned with a zero set, so it should hold for any measurable set.

    To get the result to hold on an open interval, separate f into its positive and negative parts, and you should get:

    [tex]m(F((x,y)) \leq \int _x^yf^+(t)dt + \int _x^yf^-(t)dt = |F^+(y) - F^+(x)| + |F^-(y) - F^-(x)| \leq M^+|x-y| + M^-|x-y| = (M^+ + M^-)m((x,y))[/tex]
     
  8. Dec 23, 2006 #7

    AKG

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    To clarify, the positive part of f is f+, which is defined as follows: f+(x) = f(x) when f(x) is positive, and f+(x) = 0 when f(x) is negative. [itex]F^+(x) = \int _0 ^x f^+(t)dt[/itex]. Clearly, the positive and negative parts of f are bounded and measurable, so part i applies to them, and it is by applying part i to these to functions that we get M+ and M- (you might have been wondering where those numbers came from).

    Note, the first inequality in the last line might need some explanation, but you should be able to get it for yourself. Unfortunately, I don't have time now to elaborate.
     
  9. Dec 23, 2006 #8
    Thanks so much! I think I can go from here. Actually, I thought about it on my way out and I figured it out. Thanks so much everyone!
     
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