Fubini's Theorem

eric3353 · **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
Fubinis theorem, there exists z in R^m (indeed there is a positive Lebesgue measure of such
zs) such that the line defined 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

Shadow · **Posted:** Wed, 27 Jun 2012 17:16:37 UTC

eric3353 wrote:

Im reading a paper and have come across a statement: "B must have positive Lebesgue measure in Rm: Consequently, by
Fubinis theorem, there exists z in R^m (indeed there is a positive Lebesgue measure of such
zs) such that the line defined 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 $\mathbb{R}^m$ 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 $\mathbb{R}^m$ 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.

eric3353 · **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

Shadow · **Posted:** Wed, 27 Jun 2012 17:43:21 UTC

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 $B\subseteq [0,1]^m$ 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 m-1

coordinates however you like. Then $$\int_{\mathbb{R}^m} 1_Bd\lambda=\lambda(B)>0$ where here $\lambda$ is the usual Lebesgue measure on $\mathbb{R}^m$ , then by Fubini you can break this into the multiple integral $$\int_{\ell^\perp}\int_{\ell}1_Bd\mu\,d\nu$ where $\mu$ is one-dimensional Lebesgue measure on your line, which I have named $\ell$ , and $\nu$ is the (m-1)

-dimensional Lebesgue measure on the orthogonal compliment of $\ell$ in $\mathbb{R}^m$ . But then this is equal to a positive number, and 1_B

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 $\lambda(B)>0$ . However, the inner integral is exactly $\mu(\ell\cap B)$ , so you have proven that $\mu(\ell\cap B)>0$ as desired.

eric3353 · **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...

Shadow · **Posted:** Wed, 27 Jun 2012 17:54:29 UTC

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 $\lambda$ is regular, so [0,1]^m

, being compact, has finite measure.

eric3353 · **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.

Shadow · **Posted:** Wed, 27 Jun 2012 20:00:09 UTC

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.

eric3353 · **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.

Shadow · **Posted:** Wed, 27 Jun 2012 21:36:36 UTC

Well that's certainly true, just pick something like $\ell: \mathbf{x_0}+t(1,1,1,\ldots, 1)$ where $\mathbf{x_0}\in B$ , however you're using Lebesgue measure, so points themselves have measure zero, so you need to be careful how you state things.

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