# S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
 It is currently Thu, 23 May 2013 13:03:06 UTC

 All times are UTC [ DST ]

 Page 1 of 1 [ 10 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: Fubini's TheoremPosted: Wed, 27 Jun 2012 16:57:22 UTC
 Senior Member

Joined: Wed, 28 May 2008 23:03:08 UTC
Posts: 111
Im reading a paper and have come across a statement: "B must have positive Lebesgue measure in Rm: Consequently, by
Fubinis theorem, there exists z in R^m (indeed there is a positive Lebesgue measure of such
zs) such that the line defined by z+R((1+ i); (1+ i)^2;....., (1+ i)^m) intersects B in a set of
positive one-dimensional Lebesgue measure on the line.", where the set B was a subset of the cube [0,1]^m.

Now, every time I type in "Fubini's theorem" into google, i get stuff about product spaces/multiple integrals, etc., which is what I'd expect. Is there another "fubinis theorem", possibly called something different, or a reference somebody could give me? Furthermore, does this generalize to arbitrary cubes [a_1,b_1]X....X[a_n,b_n]. Thanks a lot for you help, Ive literally been searching for this "Fubinis theorem" for two days now.

Eric

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 17:16:37 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
eric3353 wrote:
Im reading a paper and have come across a statement: "B must have positive Lebesgue measure in Rm: Consequently, by
Fubinis theorem, there exists z in R^m (indeed there is a positive Lebesgue measure of such
zs) such that the line defined by z+R((1+ i); (1+ i)^2;....., (1+ i)^m) intersects B in a set of
positive one-dimensional Lebesgue measure on the line.", where the set B was a subset of the cube [0,1]^m.

Now, every time I type in "Fubini's theorem" into google, i get stuff about product spaces/multiple integrals, etc., which is what I'd expect. Is there another "fubinis theorem", possibly called something different, or a reference somebody could give me? Furthermore, does this generalize to arbitrary cubes [a_1,b_1]X....X[a_n,b_n]. Thanks a lot for you help, Ive literally been searching for this "Fubinis theorem" for two days now.

Eric

There is only the Fubini's theorem listed in google. The idea of this is that the integral over can (under the appropriate conditions) be broken into iterated integrals over a subspace. The main drive of it is because of the fact that the Lebesgue measure on is the product measure of distinct 1-dimensional Lebesgue measures (and is the only one up to scaling by the Caratheodory extension theorem, or by uniqueness--again up to scaling--of Haar measure if you know that theorem). Also, if you're learning about Lebesgue measure, you should have seen Fubini's theorem years ago in vector calculus, perhaps review your notes from then to get the basic version down.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 17:32:41 UTC
 Senior Member

Joined: Wed, 28 May 2008 23:03:08 UTC
Posts: 111
Thank you for youre response. I'm well aware of Fubinis theorem regarding multiple integrals. I don't see how that applies to the statement that I put in my original message. Maybe it's an obvious consequence, but I just don't see it. If you do, could you point me in the right direction? thanks,

eric

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 17:43:21 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
eric3353 wrote:
Thank you for youre response. I'm well aware of Fubinis theorem regarding multiple integrals. I don't see how that applies to the statement that I put in my original message. Maybe it's an obvious consequence, but I just don't see it. If you do, could you point me in the right direction? thanks,

eric

The idea is to break the integral of the characteristic function of into an iterated integral. Since is measurable, you have that its characteristic function is integrable, and so Fubini's theorem applies. But then you can choose whatever coordinates you want, so choose that line as your first coordinate and choose the other coordinates however you like. Then where here is the usual Lebesgue measure on , then by Fubini you can break this into the multiple integral where is one-dimensional Lebesgue measure on your line, which I have named , and is the -dimensional Lebesgue measure on the orthogonal compliment of in . But then this is equal to a positive number, and is an a.e. nonnegative function, so the inner integral is nonnegative. But it cannot be zero as, if it were, then the whole integral would be zero, contradicting . However, the inner integral is exactly , so you have proven that as desired.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 17:50:42 UTC
 Senior Member

Joined: Wed, 28 May 2008 23:03:08 UTC
Posts: 111
Thanks a lot! I think I see the basics of it now, Ill work through it. In this way, it does seem that it should work for arbitrary cubes, unless I'm mistaken...

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 17:54:29 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
eric3353 wrote:
Thanks a lot! I think I see the basics of it now, Ill work through it. In this way, it does seem that it should work for arbitrary cubes, unless I'm mistaken...

Any set with finite, positive measure will do. We used positivity to make our conclusion, and finiteness to make sure it was legitimate to use Fubini in the first place. The finiteness came from monotonoicity of measures and the fact that is regular, so , being compact, has finite measure.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 19:56:31 UTC
 Senior Member

Joined: Wed, 28 May 2008 23:03:08 UTC
Posts: 111
Ok, I think I see it now. I cant do the math symbols, but assume that a set B in R^2 has positive, finite measure, call it m(B)>0. Express m(B) as the double integral of the indicator function for B. Because this can't be zero, the integrand, which is the integral over R^1, cannot be zero for y almost everywhere. Choose two of those y's, y_1 and y_2. Then since the integral with respect that those ys is not zero, there exists an x_1 such that the indicator function expressed at (x_1,y_1) is not zero, and the same for (x_2, y_2). Taking these two pairs, draw a line through them, and we're done.

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 20:00:09 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
eric3353 wrote:
Ok, I think I see it now. I cant do the math symbols, but assume that a set B in R^2 has positive, finite measure, call it m(B)>0. Express m(B) as the double integral of the indicator function for B. Because this can't be zero, the integrand, which is the integral over R^1, cannot be zero for y almost everywhere. Choose two of those y's, y_1 and y_2. Then since the integral with respect that those ys is not zero, there exists an x_1 such that the indicator function expressed at (x_1,y_1) is not zero, and the same for (x_2, y_2). Taking these two pairs, draw a line through them, and we're done.

Huh? The indicator is nonzero exactly on the intersection. Saying there are two points where the indicator is positive does not mean anything. Also, what do you mean the "integral with respect that those ys is not zero", that neither makes sense or seems correct.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 21:28:10 UTC
 Senior Member

Joined: Wed, 28 May 2008 23:03:08 UTC
Posts: 111
What I'm essentially trying to show is that if B is any set of positive finite measure in R^n, there exist two point x and x' in B such that x' is strictly componentwise larger than x. Im trying to do this by showing that there is an increasing line that must intersect B at a point where B has non-zero measure, which would give the result.

Top

 Post subject: Re: Fubini's TheoremPosted: Wed, 27 Jun 2012 21:36:36 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
Well that's certainly true, just pick something like where , however you're using Lebesgue measure, so points themselves have measure zero, so you need to be careful how you state things.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 10 posts ]

 All times are UTC [ DST ]

#### Who is online

Users browsing this forum: No registered users

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forum

Search for:
 Jump to:  Select a forum ------------------ High School and College Mathematics    Algebra    Geometry and Trigonometry    Calculus    Matrix Algebra    Differential Equations    Probability and Statistics    Proposed Problems Applications    Physics, Chemistry, Engineering, etc.    Computer Science    Math for Business and Economics Advanced Mathematics    Foundations    Algebra and Number Theory    Analysis and Topology    Applied Mathematics    Other Topics in Advanced Mathematics Other Topics    Administrator Announcements    Comments and Suggestions for S.O.S. Math    Posting Math Formulas with LaTeX    Miscellaneous