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Homework Help: Measure theory question

  1. May 7, 2010 #1
    This is a practice final exam problem that has been giving me fits: Let [itex]E[/itex] be a Lebesgue measurable subset of the interval [itex][0,1][/itex] that has finite measure. Show that there exist two points [itex]x,y \in E[/itex] such that [itex]x-y[/itex] is rational.
     
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  3. May 8, 2010 #2

    berkeman

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    Moved to HH. Could you please post your work so far? Thanks.
     
  4. May 8, 2010 #3
    Well, I haven't done much, as I'm rather stumped here. My professor gave me the following hint: "countable unions." He also mentioned that examining the non-measurable set constructed with the axiom of choice might help. But this sort of has me even more baffled.
     
  5. May 8, 2010 #4

    jbunniii

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    Why don't you start by summarizing for us the construction of the non-measurable set you're referring to? If it's the usual construction in most textbooks, it should start by defining an equivalence relation on the unit interval, such that two points [itex]x[/itex] and [itex]y[/itex] are equivalent if and only if [itex]x - y[/itex] is rational. That seems like a promising start.
     
  6. May 8, 2010 #5
    That is exactly it. We consider an equivalence relation on [itex][0,1][/itex] defined as follows: [itex]x\sim y[/itex] if [itex]x - y \in \mathbb Q[/itex]. We then employ the Axiom of Choice and choose one member [itex]x_\alpha[/itex] from each equivalence class [itex]E_\alpha[/itex], and define the set [itex] N = \{x_\alpha \}[/itex]. The uncountability of [itex]N[/itex] follows from the fact that [itex]N[/itex] must satisfy [itex]1 \leq \sum_{k=1}^\infty m(N) \leq 3[/itex] (where [itex]m[/itex] denotes Lebesgue measure).
     
    Last edited: May 8, 2010
  7. May 8, 2010 #6

    jbunniii

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    OK, now let's consider your set [itex]E[/itex]. Suppose the assertion were not true. Then there do not exist two distinct points [itex]x,y \in E[/itex] such that [itex]x - y \in \mathbb{Q}[/itex].

    Now consider the implication in terms of the equivalence classes [itex]E_\alpha[/itex]. What is the maximum number of elements of [itex]E[/itex] that can be contained in each class [itex]E_\alpha[/itex]?
     
  8. May 8, 2010 #7

    jbunniii

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    P.S. I am assuming that in the problem statement, "finite measure" should really be "finite NON-ZERO measure." Otherwise the statement is false, as the example [itex]E = \{0, 1/\sqrt{2}\}[/itex] demonstrates.
     
  9. May 8, 2010 #8
    You're right. Actually, instead of finite, I should have said "positive" (which I'm assuming means [itex]0 < mE < \infty[/itex]).
     
  10. May 8, 2010 #9
    (1) You're very kind to help me out with this. Thanks. I'd buy you a beer if I could. (Or a glass of wine; whichever floats your boat.)

    (2) Ah, I think I see...so each equivalence class consists of a single point: Given [itex]x,y\in E[/itex], [itex]x \sim y[/itex] iff [itex]x = y[/itex]. I guess this means there must be uncountably many equivalence classes, then, since [itex]mE > 0[/itex] implies it's necessarily uncountable. More importantly, don't we have [itex]N = \{ x_\alpha \} = E[/itex]? I guess we can derive a contradiction from this?
     
    Last edited: May 8, 2010
  11. May 8, 2010 #10

    jbunniii

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    Yes, "positive" makes more sense than "finite." In fact, [itex]m(E) \leq m([0,1]) = 1[/itex] so finiteness is automatic.

    OK, so far we have established that each equivalence class contains at most one point of [itex]E[/itex].

    Now we can use the axiom of choice to construct a set [itex]S[/itex] containing one point of [itex]E \cap E_\alpha[/itex] for each equivalence class [itex]E_\alpha[/itex] for which [itex]E \cap E_\alpha[/itex] is non-empty. Since each [itex]E_\alpha[/itex] contains at most one point of [itex]E[/itex], what can you say about [itex]S[/itex]?
     
  12. May 8, 2010 #11

    jbunniii

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    P.S. I love Belgian ales.:smile:
     
  13. May 8, 2010 #12
    Ok. I think my argument above has shown that each equivalence class consists of *exactly* one point, and that if we apply the Axiom of Choice (hehe) and pluck one point out of each equivalence class, at the end of the day we get [itex]N = E[/itex]. So, if we define the translates [itex]N_k = N + r_k = E + r_k[/itex] for an enumeration of the rationals in [itex][-1,1][/itex], we find that

    [tex]
    mE \leq \sum_{k = 1}^\infty mN_k \leq 3,
    [/tex]

    since it can be shown that the sets N_k are disjoint, [itex]E \subset \bigcup_{k = 1}^\infty N_k[/itex] (just take [itex]r_k = 0[/itex]) and [itex]\bigcup_{k = 1}^\infty N_k \subset [-1,2][/itex]. But this implies, by translation invariance of [itex]m[/itex], that

    [tex]
    \sum_{k=1}^\infty mE \leq 3.
    [/tex]

    This is absurd, since we assumed [itex]mE > 0 \Rightarrow \sum_1^\infty mE = \infty[/itex]. Hence [itex]E[/itex] is not measurable. Have I missed anything?
     
    Last edited: May 8, 2010
  14. May 8, 2010 #13
    La Chouffe, maybe? Chimay?
     
  15. May 8, 2010 #14

    jbunniii

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    Replying to your updated message:

    Let's be sure we agree about everything so far. Using your notation:

    The equivalence classes are called [itex]E_\alpha[/itex].

    Now each [itex]E_\alpha \cap E[/itex] contains either zero or one element. For convenience, define

    [tex]A = \{\alpha : E_\alpha \cap E \neq \emptyset\}[/tex]

    Using the axiom of choice, for each [itex]\alpha \in A[/itex] we choose [itex]x_\alpha \in E_\alpha \cap E[/itex]. We thereby construct this set:

    [tex]N = \bigcup_{\alpha \in A} \{x_\alpha\}[/tex].

    As you pointed out, we actually have [itex]N = E[/itex].

    Now let

    [tex]\{r_i\}_{i=1}^{\infty}[/tex]

    be an enumeration of the rationals in [itex][0,1][/itex]

    And define the following sets:

    [tex]N_i = N + r_i[/tex]

    where the addition is performed modulo 1.

    Then, exactly as in the construction of the standard non-measurable set, we have

    [tex]N_i \cap N_j = \emptyset[/tex]

    whenever [itex]i \neq j[/itex], i.e., the sets are disjoint.

    Then consider

    [tex]\bigcup_{i=1}^{\infty} N_i[/tex]

    What can you say about the measure of this set, and why is that a contradiction?

    P.S. For some reason the preview function is behaving strangely - it puts the wrong stuff in each TeX section. So I will have to fix any typos after I post the message.
     
  16. May 8, 2010 #15

    jbunniii

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    I assume you mean

    [tex]\sum_{k=1}^\infty[/tex]

    not

    [tex]\bigcup_{k = 1}^\infty[/tex]

    Yes, now you use the fact (just as in the standard construction of the nonmeasurable set) that the [itex]N_k[/itex] are pairwise disjoint, and also the translation invariance of Lebesgue measure, to obtain

    [tex]\bigcup_{k=1}^\infty N_k \subset [-1,2][/tex]

    and thus

    [tex]\sum_{k=1}^\infty m(E) \leq 3[/tex]

    That can't be true if [itex]m(E) > 0[/itex]. (Note that it COULD be true if [itex]m(E) = 0[/itex], which is why the positive hypothesis is important.)

    Thus we have a contradiction. Therefore our assumption was wrong, and [itex]E[/itex] must indeed contain distinct [itex]x,y[/itex] such that [itex]x - y \in \mathbb{Q}[/itex].
     
    Last edited: May 8, 2010
  17. May 8, 2010 #16
    jbunniii,

    Yes, I meant [itex]\sum[/itex] instead of [itex]\bigcup[/itex]. You've also pointed out to me that there is actually no contradiction present in

    [tex]
    mE \leq \sum_{k=1}^\infty mE.
    [/tex]

    The contradiction comes from the *other* inequality, which would force [itex]mE = 0[/itex], which is impossible. I have made the appropriate corrections to my posts above.

    Thanks again!
     
  18. May 8, 2010 #17

    jbunniii

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    Looks good! By the way, notice that we didn't actually need the axiom of choice in this case to construct the set [itex]N[/itex]. We could have simply said "let [itex]N = E[/itex]" and shown directly that the [itex]N_k[/itex]'s are pairwise disjoint.
     
  19. May 8, 2010 #18

    jbunniii

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