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Thread: Little problem in measure theory

  1. #1 Little problem in measure theory 
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    hi,

    here is a little problem from Royden. There was a proposition:

    14. Proposition: Let <En> be an infinite decreasing sequence of
    measurable sets, that is, a sequence with En+1 in En for each n. Let
    mE1 be finite. Then

    m(intersection Ei, i :1-->inf) = lim mEn as n--> inf
    Now, the problem is to show that the condition that mE1 is finite is a necessary condition, by giving a decreasing sequence <En> of measurable sets with (intersection En) = phi and mEn = inf for each n.

    I don't know... most of the decreasing sequences I found do intersect... and I found one that did not intersect in the end, but mEn is not inf...

    Also, there is a little detail about writting. I've been thinking that
    m(intersection Ei, i :1-->inf) is simply m( lim intersection Ei, i :1-->n as n-->inf). Is that true? because in that case, I beleive this intersection is just En since the sequence is decreasing. Then we need to find a sequence such that
    m(lim En) as n-->inf = 0 while m(En) = inf for each n, which doesn't seem to be quite reasonable...

    Thanks a lot!


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  3. #2  
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    Here is what I tried:

    Let E1 = R
    E2 = R \ (-1, 1)
    E3 = R \ (-2, 2)
    ...
    En = R \ (-n, n)

    Then I beleive that
    (intersection En as n--> inf) = phi
    And m(En) = inf for each n

    The problem is that I don't really know how that contradicts the proposition.

    The LHS in the statement will be zero since m(phi) = 0 (because in general m*(phi) = 0). But what about the RHS? m(En) is inf for each fixed n, but when we take n--> inf?

    I don't know...


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  4. #3  
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    You're exactly right. The limit of a sequence that is constantly infinity is infinity, as is the limit of an unbounded increasing sequence of reals.
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  5. #4  
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    Thanks as usual

    Well, if I have other problems in measure, I'll be posting them here. But I'll make sure I try really hard before I trouble you!
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  6. #5  
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    I'm having some trouble with proving that the generalized Cantor set (Cantor set with +ve measure) is nowhere dense.

    I was able to prove that the usual ternary Cantor set F is nowhere dense: suppose not and let x be in int(closure of F). Then, noting that F is closed, x is in int(F). Hence, we have a delta > 0 such that S = ]x-delta, x+delta[ is inside F. But then m(S) < or = m(F). Since m(S) = 2 delta while m(F) = 0, we have a contradiction.

    I beleive my argument can be extended to state that "if A is a closed set with measure zero, then A is nowhere dense".

    For the Cantor ternary set, I also found a proof on the net using ternary expansion.

    But what about the generalized Cantor set? I can't generalize my argument there...
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  7. #6  
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    First, you're right about the generalization of your argument to closed null sets.

    Next, I have an idea for an argument. I want you to be able to reason it out yourself, so I'm not going to give it to you straight up. But the basic idea is, again, we want to show that there are no open subintervals of your Cantor set C.

    First, however, I'd like to be able to describe the elements of C. We can write C as the intersection of a sequence of nested sets C<sub>n</sub>, each of which is a finite disjoint union of closed intervals. According to the specific details of C's construction, we know that the endpoints of these intervals are elements of C, and since C is closed, the limit points of these endpoints are also in C. Are there any other points in C?
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  8. #7  
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    yes there are, though I can't really "see" them.

    I read a proof stating that the Cantor ternary set was uncountable. This is obviously larger than the set of rational endpoints of open intervals in [0,1] which is countable.

    And since the Cantor set is a subset of the generalized cantor set (let's call the first one CS and the other one GCS), then GCS is also uncountable. Note that CS is a subset of GCS, since in GCS the intervals we remove get smaller and smaller. In the one in my book, we keep removing intervals of lenths p3^-n, where 0<p<1.
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  9. #8  
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    Okay, I've got you going down the wrong train of thought, I just realized. Let's start over.

    So C can be written as the intersection of a nested sequence of sets C<sub>n</sub> each being a finite disjoint unions of closed intervals. Let U be an open subinterval of C. Then U is an open subinterval of C<sub>n</sub>. Thinking about the description of C<sub>n</sub>, what can you say about U?
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  10. #9  
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    I can say that U is an open subinterval of one of the closed interval that unite to form Cn
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  11. #10  
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    Exactly. Now do you see where I'm going with this? What can we say about the closed subintervals as n goes to infinity?
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  12. #11  
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    I understand, but the problem is that C is a countable union of sets, hence these sets cannot be countable, otherwise C would be countable.


    It follows that the closed intervals say [i, j] cannot consist of just {i} and {j} as n-->inf. i.e., there must be points other than the end points, otherwise C would be countable. (Note that these end points are rationals).

    So, I don't see why we can't force an open interval in one of them... since we have more than the endpoints, we have some "continuity" in the intervals (which is why C has the power of continuum), so why can't we put an open interval?
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  13. #12  
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    True, if you intersect a closed interval from C<sub>n</sub> with C, you'll get more than just the endpoints--you'll get the endpoints of some of the subintervals of C<sub>m</sub>, m > n, and the limit points of such. But that's not my point--my old argument required this fact, but my new one doesn't. In fact, my new argument doesn't even look at points, just sets.

    So we know that, for each n, the open interval U is contained in some closed subinterval I<sub>n</sub> in C<sub>n</sub>. What can you say about I<sub>n</sub> as n goes to infinity?
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  14. #13  
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    I can say that In gets very small... most probably it just becomes a point, so that it can't have an open subinterval U.

    Hum... I thought about that, but the fact that C was uncountable made me feel uncomfortable... if each In becomes a point ultimately, then how can we explain that C is uncountable. Because these points are all rationals. That's why I thought I was wrong, and that each In gets very small, but not a point, so that we can force a U in it.

    But the key is probably in what you said

    and the limit points of such
    mmm, which limit points exactly? you mean the limit of the sequence of the endpoints? that would make sense, because in this case, a sequence of rationals can indeed tend towards an irrational point, so that indeed we do get irrationals and may expect the set to be uncountable.

    Ok, I beleive the problem is solved now. Hum, so the same reasoning can work for any cantor set, whether it has zero or positive measure.

    Well, thanks a lot
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  15. #14  
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    You've got the right idea! For any interval I = [a,b], let |I| = b-a. Then |I<sub>n</sub>| goes to 0 as n goes to infinity. But clearly m(U) ≤ |I<sub>n</sub>| for any n, and so m(U) = 0, whence U must be empty, a contradiction.

    We know that fat Cantor sets have to be uncountable, since they have positive measure, and all countable sets are null sets. But you can also show directly that any Cantor set is uncountable: given any sequence of nested closed subintervals I<sub>n</sub> of C<sub>n</sub>, their intersection must be a nonempty closed set (due to the compactness of [0,1]), and it must have length 0, so it must be a singleton. In fact, there is a one-to-one correspondence between points of C and sequences of nested subintervals as described above. There is an injection of the set of sequences of 0's and 1's into the set of sequences of nested subintervals--if you have a 0 in the first spot, choose the leftmost subinterval of C<sub>1</sub>, and choose the rightmost if you have a 1 in the first spot; continue in this fashion, i.e. if you're sitting in a subinterval of C<sub>n</sub>, then choose the leftmost subinterval of this interval in C<sub>n+1</sub> if you've got a 0 in the n+1st spot and the rightmost if you've got a 1. The set of sequences of 0's and 1's is uncountable (there's a surjection of this set to [0,1] given by mapping a sequence to a binary expansion), and so the Cantor set must be uncountable. (This is likely not the shortest way to show this...)

    You get a boundary point by choosing a sequence that is eventually all 0's or all 1's. You get the "other" points by choosing sequences that don't stabilize.
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  16. #15  
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    I like the proof.

    But I didn't understand your last statement

    You get a boundary point by choosing a sequence that is eventually all 0's or all 1's. You get the "other" points by choosing sequences that don't stabilize.
    We know that C is closed and nowhere dense, so can't we say it is equal to its boundary? so what are the "other points"?

    Maybe by boundary point, you mean the endpoints of the closed In? and the other points are the irrationals?
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  17. #16  
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    Yeah, the boundary points I'm talking about are the endpoints of the intervals I<sub>n</sub>, and they're also the boundary points in the topological sense of the sets C<sub>n</sub>.
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  18. #17  
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    hum I see.

    Sorry to trouble you, but why does this choice (eventually all 0 or all 1) gives a boundary point while the other gives the other kind of points?

    mmm if I have a 0 I choose the leftmost subinterval, if I have a 1 I choose the rightmost... so having (smth)00000...means I keep choosing left left left... probably I reach what would be the left endpoint of a closed interval? and in the (smth)1111 I reach the right endpoint, while the unstable one gives me a point in between... that's of course when we're still with n then take n--> inf...

    Am I correct?
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  19. #18  
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    Indeed. And you can never get to an endpoint if you don't keep choosing the interval it's in (as you'll be separated from it by an open interval). Since an endpoint is always going to be the left or right endpoint, i.e. it never switches sides, you've always got to be choosing the side that the endpoint is on.
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  20. #19  
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    yes, it makes sense now.

    well thanks very much
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  21. #20  
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    hi, it has been a long time

    a little question about Fatou's lemma: I'll denote by "int (fn)" the Lebesegue integral of fn and by lim inf the limit infimum of fn


    if we have a sqn {fn} of nonnegative measurable fns with fn -->f, then int(f) < or = lim inf int(fn)
    I understand the proof in my book, but I think he complicates things.

    Here is my alternative proof:
    inf fn < or = fn for all n
    ==> sup inf fn < or = fn for all n
    ==> lim inf fn < or = fn for all n
    ==> int (lim inf fn) < or = int (fn) for all n
    ==> int (lim inf fn) < or = lim inf int (fn)
    But fn -->f ==> lim inf fn = lim sup fn = f
    ==> int(f) < or = lim inf int(fn)

    There should be a mistake because this proof is much simpler, but I really cannot locate it. I used the fact that the sequence is nonnegative in the third implication (about the integral). So, is there anything wrong?

    Thanks very much
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  22. #21  
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    Nevermind what I just wrote; it was completely dumb. I just found the terrible mistake.

    It's quite a shame I can still write that kind of proofs... whatever...

    But I was a bit nostalgic about the forum, so it's not that bad I hope you're doing well.

    Thanks and I'll try to be more careful
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  23. #22  
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    You've gotta remember the indices on the sups and infs when you're defining liminf. Limit infimum is the supremum over m of the infimum over n greater than m. So the line:

    sup inf f<sub>n</sub> ≤ f<sub>n</sub> for all n

    does not work. In fact, let's come up with a counterexample. Let f<sub>n</sub> be the characteristic function of the interval [1/n,1]. Then f<sub>n</sub> converges to f, the characteristic function of (0,1], and hence we have f = liminf f<sub>n</sub>. But for no value of n is it true that f ≤ f<sub>n</sub>.
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  24. #23  
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    yes, it's about the "sup inf" part.

    in fact, using my reasoning, you can prove brilliant results like "any increasing sequence is divergent". This comes as a corollary of "lim inf fn < or = fn for all n". lol.
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  25. #24  
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    hi there

    I was trying to solve a problem here. I wrote a solution, but there is a little step which I'm not sure is "legal". Here we are: it's a problem in Royden real analysis, problem 4.17b :

    Let f be integrable over (-inf, inf) and g be a bounded measurable function. Then

    lim t-->0 {int(- inf, inf){|g(x)[f(x+t) - f(x)]|}} = 0
    Here is my proof:
    I'll first work the case where f is uniformly continuous on a finite interval. The generalzition should be easy from a previous problem. So assume f is u.c and let eps>0. Then there is a delta>0 such that |x - (x+t)| < delta ==> |f(x+t) - f(x)|< eps
    Hence, |t| < delta ==> |f(x+t) - f(x)|< eps
    i.e.
    lim t-->0 {|f(x+t) - f(x)|} = 0
    Since g is a bounded function, then
    lim t-->0 {|g(x)[f(x+t) - f(x)]|} = 0
    Hence,
    int(-inf, inf) {lim t-->0 {|g(x)[f(x+t) - f(x)]|}} = 0

    Everything is alright till now. The trick now is to puch that integral in the limit. So here is how I reasoned: what we have here is almost the integration of a sequence, but instead of an integer variable n, we have a real variable t. So the trick would be to change that sequence as follows: let t = 1/n. Then

    int(-inf, inf) {lim n-->inf {|g(x)[f(x+1/n) - f(x)]|}} = 0

    Now let M>0 and assume |g(x)| < M . Then we have {|g(x)[f(x+1/n) - f(x)]|}} < M |f(x)| which is integrable

    Then from Lebesgue convergence theorem,

    lim n-->inf {int(-inf, inf) {|g(x)[f(x+1/n) - f(x)]|}} = 0

    Thus,

    lim t-->0 {int(- inf, inf){|g(x)[f(x+t) - f(x)]|}} = 0

    The only step I'm not sure about, is whether I can change t by 1/n, infer some properties, then go back to t. Is that ok?

    Thanks very much
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  26. #25  
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    I just came across an interesting problem: Problem 2.49f:

    lim x-->y {f(x)} = L <==> for every sequence {xn} such that xn is not equal to y, xn --> y , lim n--> inf {f(xn)} = L
    So here is the improvement I'll make for my proof: instead of taking t = 1/n, let {tn} be any sequence such that tn --> 0. With that, I beleive my proof is flawless.

    Thanks very much anyway!
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