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Lebesgue measure on R fact

  1. Dec 16, 2007 #1
    Just thought for food (not exactly a homework question).

    Find a Borel subset E of [0,1] such that for any subinterval I of [0,1],
    0 < m(E [itex]\cap[/itex] I) < m(I).

    If you know the answer, post it. Otherwise, I'll post it when I figure it out.
  2. jcsd
  3. Dec 16, 2007 #2
    I found the sketch to the answer to this problem in Makarov (and others) Selected Problems in Real Analysis. Here is what it looks like.

    This first requires a lemma (proved last):
    For any open interval (a,b), and any e such that 0 < e < b - a, there exists a compact totally disconnected subset K of (a,b) such that m(K) = e. (Totally disconnected means K contains no interval with more than one point.)

    It suffices to consider (0,1) instead of [0,1] in the above problem statement. First construct a totally disconnected compact [itex]K \subset (0,1)[/itex] such that m(K) = 1/4. Then (0,1) \ K is open, hence we can set [tex]U_{1,1}, U_{1,2}, U_{1,3}, ...[/tex] to be the disjoint open intervals making up (0,1) \ K. Now for each n, let [tex]K_{1,n}[/tex] be a compact totally disconnected subset of [tex]U_{1,n}[/tex] such that [tex]m(K_{1,n}) = \frac{ m(U_{1,n}) } {2^3}[/tex]. Let [tex]U_{2,1}, U_{2,2}, U_{2,3}, ...[/tex] be the open intervals making up [tex]\bigcup_{n=1}^\infty (U_{1,n} - K_{1,n})[/tex]. For all n, let [tex]K_{2,n}[/tex] be a compact totally disconnected subset of [tex]U_{2,n}[/tex] such that [tex]m(K_{2,n}) = \frac{ m(U_{2,n}) } {2^4}[/tex]. Continue constructing K's, U's inductively, with the identity [tex]m(K_{i,j}) = \frac{m(U_{i,j})}{2^{j+2}}[/tex].

    Define [tex]E = K \cup \bigcup_{i=1}^\infty \bigcup_{j=1}^\infty K_{i,j}[/tex]. We claim this is the desired Borel subset of (0,1). Namely, for any subinterval I of (0,1), we can find a [tex]U_{i,j} \subset I[/tex], and hence [tex]0 < m(K_{i,j}) \leq m(E \cap I)[/tex]. And also, ..... found error here!! (The prior conclusion is false: [tex]m(E \cap I) \leq m( K_{i,j} \cup I - U_{i,j}) = m(K_{i,j}) - m(U_{i,j}) + m(I) < m(I)[/tex].)

    Proof of lemma:

    Suffices to look at [0,1], (instead of looking at [a+d,b-d] subset (a,b) with small d > 0), and it suffices to show for any 0 < e < 1 there exists a dense open U subset (0,1) such that m(U) = e. (Then set K = [0,1] - U, to get m(K) = 1 - e.) To construct such a U, first enumerate the rationals in (0,1) as R = {x1,x2,...}.

    Choose p1 such that U1 = (x_p1-e/4,x_p1+e/4) is a subset of (0,1). Then let R1 = R - cl(U1). Choose p2 such that x_p2 in R1, and U2 = (x_p2-e/8, x_p2 + e/8) is a subset of (0,1) and does not intersect U1. Let R2 = R1 - cl(U2). Continue this process, and you get [tex]U = \bigcup_{i=1}^\infty U_i[/tex] is open, and is a disjoint collection such that every rational is in some Ui or otherwise is an endpoint, hence is dense, and we have [tex]m(\bigcup_{i=1}^\infty U_i) = \sum_{i=1}^\infty m(U_i) = \sum_{i=1}^\infty \frac{e}{2^i} = e[/tex].
    Last edited: Dec 16, 2007
  4. Dec 17, 2007 #3
    Would have just edited the above, but there were a lot of errors. In the lemma, we also need to, if necessary, add a finite number of points to K to ensure the open intervals in (0,1) - K have at length at most half of (0,1). You need to require this, so that you will be ensured for any open interval I, there will definitely be some U_ij [tex]\subset[/tex] I. And that is the main point of the proof that 0 < m(K_ij) <= m(E [itex]\bigcap[/itex] I).

    Also notice that the measure of E in any U_ij is [tex]m(E \cap U_{i,j}) = \sum_{i=2}^\infty \frac{m(U_{i,j})}{2^i} = \frac{m(U_{i,j})}{2}[/tex]. Thus [tex]m(I) = m(I-U_{i,j}) + m(U_{i,j}) > m(I-U_{i,j}) + \frac{m(U_{i,j})}{2} = m(I-U_{i,j}) + m(E \cap U_{i,j}) \geq m(E \cap I)[/tex].

    I think that should fix it.
  5. Dec 27, 2007 #4
    I do see a problem in the proof of the lemma.

    How do we know that we can always assign such an open interval?

    For example, let's say e=1 (or .99999) and we've assigned open intervals of 1/2 and 1/4 like this:

    Code (Text):

    [-----(               )-----(                              )-----]

    0    1/12           4/12   5/12                          11/12   1
    Now we need to assign 1/8 somewhere, but the 3 available gaps are each only 1/12.

    So this is how I would prove the lemma:

    Enumerate the rationals [tex]{r_1,r_2,r_3,\ldots[/tex] To each rational [tex]r_{i}[/tex] assign the number [tex]e_i = {e\over{2^i}}[/tex]. Now construct U by unioning each [tex]U_i[/tex], where each [tex]U_i[/tex] is an open interval around [tex]r_i[/tex] of size [tex]e_i[/tex], unless that open interval would intersect part of U that's already been constructed, in which case we take the amount of the overlap and just assign it somewhere else into some gap(s).

    In my example, let's say we're now assigning an interval of size 1/8. If the matching rational falls into one of the three gaps, we fill in the gap completely, otherwise just pick one gap at random and fill it in. We've now assigned 1/12. Take the remaining 1/8 - 1/12 = 1/24 and assign it to one of the other two gaps, and we now get something like this:

    Code (Text):

    [--(                                                       )-----]

    0   1/24                                                11/12   1
    Keep doing this, and in the limit we end up with an open set of measure e that includes all the rationals (the answer to #7). Taking the complement (like in the lemma) gives us the answer to #6 - a totally disconnected compact set with some specified measure. Now we can proceed like in the original post to get the answer to #8, which I did a little differently.

    For those without the book, all of this is from page 58 in Rudin's Real and Complex Analysis. The problems are (almost) the same.
  6. Dec 27, 2007 #5
    As a thanks for your correction, I will try to prove again elaborating on another shady point... The other point the proof was hinging on was also totally wrong, namely [tex]m(E \cap U_{i,j}) = \sum_{i=2}^\infty \frac{m(U_{i,j})}{2^i} = \frac{m(U_{i,j})}{2}[/tex].

    I'm going to need to fix this...
  7. Dec 27, 2007 #6
    To summarize the lemma: Given any interval (a,b), (a,b], [a,b), [a,b], then for any 0 < e <= 1 there exists:
    (1) a compact totally disconnected subset K such that m(K) = (1-e)(b-a)
    (2) an open dense subset U such that m(U) = e(b-a)

    Claim: There exists a Borel subset E of [0,1] such that 0 < m(E [itex]\cap[/itex] I) < m(I) for any subinterval I of [0,1].


    Again, look at U_0 = (0,1) without loss, and fix 0 < e < 1. Let K_0 be a compact totally disconnected subset of U_0 such that m(K_0) = e/2, and add points if necessary to ensure it's complement contains no interval with size larger than 1/2. Let U_1 = U_0 - K_0. Then U_1 is open so is a countable union of disjoint open segments U_11, U_12, U_13,... For each j let K_1j be a compact totally disconnected subset of U_1j such that [tex]m(K_{1,j}) = e\frac{m(U_{1,j})}{2^2}[/tex], and such that U_1j - K_1j does not have an interval of size larger than 1/4. Let [tex]K_1 = \cup_{j=1}^\infty K_{1,j}[/tex]. Then K_1 is a Borel set that is not necessarily closed (or compact), but we still have U_1 - K_1 is open (check with simple set theory). And by countable additivity, we have [tex]m(K_1) = \sum_{j=1}^\infty m(K_{1,j}) = \frac{e}{2^2} \sum_{j=1}^\infty m(U_{1,j}) = \frac{e}{2^2} m(U_1)[/tex] .

    Now inductively, given U_n and K_n have been so defined (as above with U_n-K_n open), define [tex]U_{n+1} = U_n - K_n[/tex]. Let [tex]U_{n+1,1}, U_{n+1,2}, ...[/tex] be the disjoint open sets making up [tex]U_{n+1}[/tex]. Then let [tex]K_{n+1,j}[/tex] be a compact totally disconnected subset of [tex]U_{n+1,j}[/tex] such that [tex]m(K_{n+1,j}) = e \frac{m(U_{n+1,j})}{2^{n+2}}[/tex] and such that [tex]U_{n+1,j} - K_{n+1,j}[/tex] has no interval of measure larger than [tex]\frac{1}{2^{n+2}}[/tex]. Then let [tex]K_{n+1} = \cup_{j=1}^\infty K_{n+1,j}[/tex], and it then follows that [tex]K_{n+1}[/tex] is a Borel set and [tex]U_{n+1} - K_{n+1}[/tex] is open, and [tex]m(K_{n+1}) = \sum_{j=1}^\infty m(K_{n+1,j}) = \frac{e}{2^{n+2}} \sum_{j=1}^\infty m(U_{n+1,j}) = \frac{e}{2^{n+2}} m(U_{n+1})[/tex].

    After defining the K_n's, U_n's inductively, define [tex]E = \cup_{j = 0}^\infty K_j[/tex], [tex]U = \cap_{j = 0}^\infty U_j[/tex]. The explicit value for m(E) (or m(U)) is not exactly clear (because the definition of the size of the m(K_ij)'s shrinks as m(U_i)'s shrink), but it's clear that [tex]0 < m(K_0) < m(E) = \sum_{n=0}^\infty m(E_n) < \sum_{n=1}^\infty \frac{e}{2^{n+1}} = e[/tex]. The crucial point in the proof is that for any U_ij as defined above, we have 0 < m(E [itex]\cap[/itex] U_ij) < m(U_ij). To see this, First note K_r [itex]\cap[/itex] U_ij = empty for 0 <= r < i, and E_i [itex]\cap[/itex] U_ij = K_ij. By analogy of U_ij = U_0, K_ij = C_0, and proceeding with the same construction of U_n's and C_n's (defined like the K_n's, you get C_nj = the K_np inside U_ij), you will end up with E [itex]\cap[/itex] U_ij. But the starting value is [tex]m(K_{i,j}) = \frac{e}{2^{i+1}}m(U_{i,j})[/tex], so the sum of all the K_mn's in U_ij yields [tex]0 < m(K_{i,j}) < m(E \cap U_{i,j}) = \sum_{n=0}^\infty m(C_n) < \sum_{n=1}^\infty \frac{e}{2^{i + n + 1}}m(U_{i,j}) = \frac{e}{2^{i+1}} m(U_{i,j})) < m(U_{i,j})[/tex].

    Now you can fix any subinterval I of (0,1). You can assume I is open. Baire's Theorem makes it easier to find a U_ij in I, each U_n is a dense open subset, so U (the intersection) is dense. So there is some x in U that is in I. Then you can choose n such that (x - 1/2^n, x + 1/2^n) is contained in I, then (by the construction) every open interval of U_n is of size smaller than (or equal) 1/2^n, hence there is some U_nj containing x and it's a subset of I. It then follows directly that 0 < m(E [itex]\cap[/itex] I) < m(I).
  8. Dec 27, 2007 #7
    I should have mentioned why U_n is dense for all n.

    Suppose U_n is dense, and write [tex]U_n = \cup_{j=1}^\infty U_{n,j}[/tex], then [tex]U_{n+1} = \cup_{j=1}^\infty U_{n,j} - K_{n,j}[/tex]. Then U_nj - K_nj is dense in U_nj for all j. (If A is open and B subset A is totally disconnected, then A - B is dense in A.) The point then is that if C is dense in B, and B is dense in A, then C is dense in A... That inductively shows the U_n's are dense, which I suppose is easier than showing U is dense..
  9. Dec 27, 2007 #8
    I was actually hoping for an even stronger statement:

    For all 0 < e < 1 there exists a Borel set E such that m(E [itex]\cap[/itex] I) = em(I) for every subinterval I of (0,1).

    In other words, I couldn't see a way to distribute the K_ij's uniformly on the interval as I constructed above (i.e. you could have m(E [itex]\cap[/itex] (0,1/2)) = 1/2 and m(E [itex]\cap[/itex] (1/2,1)) = 1/8 with the above type construction....)
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