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Topic: Wanna help me study for my measure theory final? (Read 889 times) |
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Hooie
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Wanna help me study for my measure theory final?
« on: Jun 13th, 2006, 2:30pm » |
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My final's tomorrow morning, so this is kind of short notice, but any help with the following problems would be appreciated.
http://www.math.ucla.edu/classes/math/06s/131c.1.06s/review.pdf
(Sorry that I have to link to them, but I don't know how to put math symbols in my post.)
The only ones I need help with are 2.i, 2.ii, and 11.
2.i:
f is a bounded function on [0,1].
o(p:N) = sup f(x) - inf f(x), taken over x in a neighborhood N.
o(p) = inf o(p:N), taken over all N.
I'm asked to show that the set of points p such that o(p:N) < c is an open set. (I think that's a typo, since o(p:N) depends on N, not on p. My prof must have meant the set of p where o(p) < c is open.)
2.ii:
Given 2.1, show that the function that maps p to o(p) is Borel measurable.
My textbook has a theorem that says a function is mu-measurable iff the set {x: f(x) < c} is mu-measurable. So, since by the previous part, {x: o(x) < c} is open, and open sets in R are unions of open intervals and thus Borel sets, I'm done, right? Is there another way to do it without the theorem?
11:
This problem has to do with the Lebesgue interal of a function over (0, infinity) existing if the Riemann integral over every interval (a,A) exists, where 0 < a < A < infinity.
There's a typo in the statement: where is says the Riemann integral of f(x)dx is < A, it should be < L.
Again, any help would be appreciated. Thanks!
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Icarus
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Re: Wanna help me study for my measure theory fina
« Reply #1 on: Jun 13th, 2006, 5:40pm » |
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2.i) I agree about the typo. Note though that o(p:N) depends only on N, not p. Let S be the set of all p with o(p) < c, and let p ( S. Then there is at least one neighborhood N of p such that o(p:N) < c. Let q ( N, then o(q:N) = o(p:N) < c, and so o(q) < c. Hence N is a subset of S, and therefore S is open.
2.ii) There are always other ways of proving things (if nothing else, it is always possible to state what is effectively the same proof but with different terminology). However, I am sure that your proof is exactly what was desired. The theorem you state is basic.
11) I have yet to look at.
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Icarus
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Re: Wanna help me study for my measure theory fina
« Reply #2 on: Jun 13th, 2006, 6:11pm » |
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For (11), I am unsure what is acceptable to use in the proof. The problem itself is a fairly basic result in measure theory, so I am unsure how far back we must go.
For example, if we are allowed to use the fact that Riemann Integrability ==> Lesbegue Integrability on finite intervals, with equal values, and are allowed to use the Monotone Convergence Theorem, then we can proceed as follows:
Define fk(x) = f(x) if 1/k < x < k, and fk(x) = 0 if x <= 1/k or x >= k, then fk converge monitonically upward to f. Since by assumption, f is Reimann integrable on [1/k, k], it is also Lesbegue integrable on the same interval, and R(f; 1/k, k) = L(f; 1/k, k) = L(fk; 0, oo) (where the "L" denote Lesbegue integration, and similarly for the "R").
Since each of the fk are L-integrable on (0, oo), are all >= 0, and converge upward to f, f itself is L-integrable on E = (0, oo), and limk L(fk; E) = L(f, E). By the previous equality, limk R(f; 1/k, k) = L(f, E).
This is almost what is asked for. We need the limit converging for all a, A, not just for 1/k, k. But note that since f >= 0, R(f; a, A) is increasing in A, and decreasing in a. By assumption R(f; a, A) is bounded above, and therefore lima,A R(f; a, A) must converge to sup{R(f; a,A) : 0 < a < A}.
Since the limit converges, limk R(f; 1/k, k) must converge to the same value.
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"Pi goes on and on and on ...
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Hooie
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Re: Wanna help me study for my measure theory fina
« Reply #3 on: Jun 13th, 2006, 8:33pm » |
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 Thanks a lot, Icarus.
Yes, we're allowed to use the equality on finite intervals and the Monotone Convergence Theorem (though we call it Lebesgue's Dominated Convergence Theorem).
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Icarus
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Re: Wanna help me study for my measure theory fina
« Reply #4 on: Jun 13th, 2006, 8:54pm » |
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Though related, the Monotone Convergence Theorem and the Dominated Convergence Theorem are not the same.
The MCT requires that the sequence be either increasing and dominated below, or else that it be decreasing and dominated above. The DCT requires that the sequence be dominated both above and below but drops the requirement that it be monotone.
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"Pi goes on and on and on ...
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I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Hooie
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Re: Wanna help me study for my measure theory fina
« Reply #5 on: Jun 13th, 2006, 9:18pm » |
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 Oops. I typed in "monotone convergence theorem" into wikipedia, and it linked me to the article for DCT. I didn't bother reading it, I just assumed they were the same.
Well, now that I've actually read the article, I'm pretty sure our name for the MCT is Beppo Levi's Theorem.
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