| Author |
Message
|
| Stranger |
 Posted: Sat Mar 08, 2008 5:15 am Post subject: Little problem in measure theory |
 |
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
hi,
here is a little problem from Royden. There was a proposition:
| Quote: |
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!
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
 |
| Stranger |
| Posted: Sat Mar 08, 2008 6:26 am Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
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...
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
| serpicojr |
| Posted: Sat Mar 08, 2008 12:11 pm Post subject: |
|
|
Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
Location: JRZ
|
| 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. |
|
| Back to top |
|
|
| Stranger |
| Posted: Sat Mar 08, 2008 2:57 pm Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
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!
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
| Stranger |
| Posted: Sat Mar 08, 2008 10:57 pm Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
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...
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
| serpicojr |
| Posted: Sun Mar 09, 2008 1:24 pm Post subject: |
|
|
Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
Location: JRZ
|
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 Cn, 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? |
|
| Back to top |
|
|
| Stranger |
| Posted: Sun Mar 09, 2008 6:15 pm Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
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.
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
| serpicojr |
| Posted: Sun Mar 09, 2008 8:18 pm Post subject: |
|
|
Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
Location: JRZ
|
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 Cn each being a finite disjoint unions of closed intervals. Let U be an open subinterval of C. Then U is an open subinterval of Cn. Thinking about the description of Cn, what can you say about U? |
|
| Back to top |
|
|
| Stranger |
| Posted: Sun Mar 09, 2008 9:57 pm Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
I can say that U is an open subinterval of one of the closed interval that unite to form Cn
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
| serpicojr |
| Posted: Sun Mar 09, 2008 10:28 pm Post subject: |
|
|
Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
Location: JRZ
|
| Exactly. Now do you see where I'm going with this? What can we say about the closed subintervals as n goes to infinity? |
|
| Back to top |
|
|
| Stranger |
| Posted: Mon Mar 10, 2008 10:11 am Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
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?
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
| serpicojr |
| Posted: Mon Mar 10, 2008 3:36 pm Post subject: |
|
|
Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
Location: JRZ
|
True, if you intersect a closed interval from Cn with C, you'll get more than just the endpoints--you'll get the endpoints of some of the subintervals of Cm, 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 In in Cn. What can you say about In as n goes to infinity? |
|
| Back to top |
|
|
| Stranger |
| Posted: Mon Mar 10, 2008 4:30 pm Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
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
| Quote: |
| 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
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
| serpicojr |
| Posted: Mon Mar 10, 2008 5:07 pm Post subject: |
|
|
Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
Location: JRZ
|
You've got the right idea! For any interval I = [a,b], let |I| = b-a. Then |In| goes to 0 as n goes to infinity. But clearly m(U) ≤ |In| 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 In of Cn, 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 C1, 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 Cn, then choose the leftmost subinterval of this interval in Cn+1 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. |
|
| Back to top |
|
|
| Stranger |
| Posted: Mon Mar 10, 2008 7:00 pm Post subject: |
|
|
Forum Sophomore
Joined: 15 Sep 2005
Posts: 121
|
I like the proof.
But I didn't understand your last statement
| Quote: |
| 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?
_________________
Watch what thy eyes can't see... and live it.
(T.B) |
|
| Back to top |
|
|
|
|
