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Math Help - Lebesgue measure of some spaces...

  1. #1
    Moo
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    Lebesgue measure of some spaces...

    Hi,

    Again, I've seen these questions in some other forums, and I wonder how it is solved...


    1.
    Let A be the set of x in [0,1] such that the decimal expansion of x^2 does not contain any 9.
    What is the Lebesgue measure of A ?

    My thoughts on it :
    The easiest answer would be 1, by proving that A is a countable set. But I don't think it's true, because in particular, 0.95² ~ 0.9
    So, A is at least of Lebesgue measure 0.05, isn't it ?



    2.
    Let (X,B,\mu) be a measured space and f ~:~ X \to \mathbb{C} a measurable function.
    Say why B=\{x \in X ~:~ |f(x)|\geq \alpha\} belongs to B.
    We assume that \mu(B)\neq 0
    Show that there exists \alpha>0 such that \mu\left(\{x \in X ~:~ |f(x)|\geq \alpha\}\right)>0


    That's the exact way it was asked, and I don't really know if it is normal that we have to prove that B belongs to B o.O Maybe it's 2 different sets. So let C=\{x \in X ~:~ |f(x)|\geq \alpha\}

    Anyway, I don't see how to find D such that f^{-1}(D)=C



    I guess I'm already rusty with measure theory
    Last edited by Moo; April 10th 2009 at 01:57 PM.
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  2. #2
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    Hi,

    Quote Originally Posted by Moo View Post
    1.
    Let A be the set of x in [0,1] such that the decimal expansion of x^2 does not contain any 9.
    What is the Lebesgue measure of A ?

    My thoughts on it :
    The easiest answer would be 1, by proving that A is a countable set. But I don't think it's true, because in particular, 0.95² ~ 0.9
    So, A is at least of Lebesgue measure 0.05, isn't it ?
    I don't understand your last sentence. True A is uncountable, but it is negligible anyway. One fact to keep in mind: (for Lebesgue measure,) almost every real number in [0,1] has every finite pattern (like "31415925") infinitely many times in its decimal expansion. In particular, almost every real number has a 9 (and even infinitely many, and arbitrarily long sequences of 9's) in its decimal expansion.

    For a proof: first consider the set B of the numbers x in [0,1] with no "9" in their decimal expansion. B is a subset of the set B_n of the numbers in [0,1] with no "9" among their n first decimals. However, we have \lambda(B_n)=\left(\frac{9}{10}\right)^n (where \lambda is Lebesgue measure), hence \lambda(B)\leq\lambda(B_n)\to_n 0.

    Then A is obtained from B in a "smooth" way (taking the square root): A=\{\sqrt{x}|x\in B\}. Here is one way to show that this implies \lambda(A)=0: \lambda(A)=\int_0^1{\bf 1}_A(u)du=\int_0^1 {\bf 1}_B(u^2)du =\int_0^1{\bf 1}_B(t)\frac{dt}{2\sqrt{t}}=0 (the function we integrate is 0 almost everywhere).

    There are more general conditions that ensure that the image f(B) of a negligible set is negligible (for instance you can prove this is true if the function is Lipschitz).

    2.
    Let (X,\mathcal{B},\mu) be a measured space and f ~:~ X \to \mathbb{C} a measurable function.
    Say why B_\alpha=\{x \in X ~:~ |f(x)|\geq \alpha\} belongs to \mathcal{B}.
    We assume that \mu(\{x\in X~:~|f(x)|>0\})\neq 0
    Show that there exists \alpha>0 such that \mu\left(\{x \in X ~:~ |f(x)|\geq \alpha\}\right)>0
    Notice I made a few corrections; notably, I guessed the assumption actually is \mu(\{x\in X~:~|f(x)|>0\})>0 (otherwise the question is in the assumption).

    For the first question, your idea is right, just notice that |f(x)|\geq \alpha means f(x)\in C_\alpha for some closed (hence measurable) subset C_\alpha\subset \mathbb{C}. So that B_\alpha=f^{-1}(C_\alpha)\in\mathcal{B}.

    For the second question, notice \{x\in X~:~|f(x)|>0\} is the union of the (increasing) sequence of sets B_{1/n}, n\geq 1...

    Au plaisir ,
    Laurent.
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  3. #3
    Moo
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    Quote Originally Posted by Laurent View Post
    Hi,

    I don't understand your last sentence.
    Yes, it's normal you don't understand it... Because I meant of "measure at most 0.95"

    True A is uncountable, but it is negligible anyway. One fact to keep in mind: (for Lebesgue measure,) almost every real number in [0,1] has every finite pattern (like "31415925") infinitely many times in its decimal expansion. In particular, almost every real number has a 9 (and even infinitely many, and arbitrarily long sequences of 9's) in its decimal expansion.
    Hmmm I have some problems understanding this

    On the contrary, we can find infinitely many numbers which decimal expansion does contain 9.
    Things are popping up in my head... we can restrict the problem to the set of irrationals, that is with no given pattern, since the set of rationals has a Lebesgue measure 0.
    And the probability (urgh... I don't know how else to say it) to get no 9 in an infinite decimal expansion is 0.
    If it's correct, I guess the intuitive way is clear to me.

    Hmmm... while reading the end, it looks like it's the correct way ><


    For a proof: first consider the set B of the numbers x in [0,1] with no "9" in their decimal expansion. B is a subset of the set B_n of the numbers in [0,1] with no "9" among their n first decimals. However, we have \lambda(B_n)=\left(\frac{9}{10}\right)^n (where \lambda is Lebesgue measure), hence \lambda(B)\leq\lambda(B_n)\to_n 0.
    Understood !

    Then A is obtained from B in a "smooth" way (taking the square root): A=\{\sqrt{x}|x\in B\}. Here is one way to show that this implies \lambda(A)=0: \lambda(A)=\int_0^1{\bf 1}_A(u)du=\int_0^1 {\bf 1}_B(u^2)du =\int_0^1{\bf 1}_B(t)\frac{dt}{2\sqrt{t}}=0 (the function we integrate is 0 almost everywhere).
    i think i got it, but how would you translate "smooth" ?

    There are more general conditions that ensure that the image f(B) of a negligible set is negligible (for instance you can prove this is true if the function is Lipschitz).
    I don't really see why :s
    can we use the topological definition of continuity ? That the reciprocal image is contained in the initial set, or something like that...

    Notice I made a few corrections; notably, I guessed the assumption actually is \mu(\{x\in X~:~|f(x)|>0\})>0 (otherwise the question is in the assumption).
    Indeed...

    For the first question, your idea is right, just notice that |f(x)|\geq \alpha means f(x)\in C_\alpha for some closed (hence measurable) subset C_\alpha\subset \mathbb{C}. So that B_\alpha=f^{-1}(C_\alpha)\in\mathcal{B}.
    But since it's > \alpha, wouldn't it be in the complement of a circle ? Thus in an open set ? o.O



    Tout le plaisir est pour moi avoir des réponses aussi détaillées est un luxe ^^
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  4. #4
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    Quote Originally Posted by Moo View Post
    Yes, it's normal you don't understand it... Because I meant of "measure at most 0.95"
    Fine, then !...

    i think i got it, but how would you translate "smooth" ?

    I don't really see why :s
    can we use the topological definition of continuity ? That the reciprocal image is contained in the initial set, or something like that...
    "smooth" could be "continuously differentiable" (of class \mathcal{C}^1), here. Here's why. First, for the Lipschiptz function:

    Suppose f:\mathbb{R}\to\mathbb{R} is Lipschitz with constant C on the interval I: |f(y)-f(x)|\leq C|y-x| for x,y\in I. Then, for any negligible subset A of I, f(A) is negligible.

    Indeed: Let \varepsilon>0. Since \lambda(A)=0, A can be covered with intervals I_i,i\in\mathbb{N} with total length \sum_i |I_i|\leq\varepsilon. Then f(A) is covered by the subsets (intervals, in fact) f(I_i), and we notice that |f([a,b])|\leq C|b-a|, hence the total length of the covering (f(I_i))_{i\in\mathbb{N}} is less than C\sum_i |I_i|\leq C\varepsilon.

    Now, if f is continuously differentiable on I, it is Lipschitz on compact subsets of I (with C=\max_{t\in K} |f'(t)| on K\subset I). Thus if I is compact, this is ok. Otherwise, we can choose an increasing sequence (K_n)_n of compact subsets of I such that \bigcup_n K_n =I (since I is an interval, this is easy), and apply the previous reasoning to A_n=A\cap K_n to show that \lambda(f(A_n))=0. Since \lambda(f(A))=\sup_n \lambda(f(A_n)), we can conclude \lambda(f(A))=0.

    For instance, in your case this result applies to f(t)=\sqrt{t} on I=(0,1). But I gave a quicker proof in this particular case, using a change of variable.

    But since it's > \alpha, wouldn't it be in the complement of a circle ? Thus in an open set ? o.O
    Well, nobody wrote " >\alpha" but if it should be this way, then you're right. Anyway, the important thing is that is measurable (i.e. borelian), which is the case.

    Laurent.
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  5. #5
    Moo
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    Hi again,

    I'm stupid ><

    The actual question (for 2.) was : say why B=\{x \in X ~:~ f(x) \neq 0\} belongs to \mathcal{B}


    I still have some difficulties in understanding your last message though, so I'll need some time


    Well, nobody wrote ">\alpha" but if it should be this way, then you're right. Anyway, the important thing is that is measurable (i.e. borelian), which is the case.
    I actually meant \geq \alpha, not > \alpha.. Again sorry for the misunderstandings!


    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    To sum up, the initial problem is :
    Let (X,\mathcal{B},\mu) be a measured space and let f : X \to \mathbb{C} is a measurable function.
    Say why B=\{x \in X ~:~ f(x) \neq 0\} belongs to \mathcal{B}

    Let's suppose that \mu(B)\neq 0. Show that there exists \alpha>0 such that \mu\left(\{x \in X ~:~ |f(x)|\geq \alpha\}\right)=0


    So for the first part, it's the same reasoning as above, \mathbb{C}^* is an open set in the complex plane (?)




    Conclusion : I really needed this vacation
    Last edited by Moo; April 17th 2009 at 11:58 AM.
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