User damek davis - MathOverflow

Answer by Damek Davis for nontrivial theorems with trivial proofs

2011-05-25T03:17:26Z

Chebyshev's inequality is the following:

Suppose $X, \mu$ is a measure space, and $f \in L^p(X, \mu)$, then for all $t > 0$

$\mu( {x \in X : |f(x)| \geq t } ) \leq \frac{1}{t^p} \|f\|_{L^p(X, \mu)}^p$.

The proof is trivial:

Observe that

$\mu( {x \in X : |f(x)| \geq t } )t^p = \int_{X} 1_{|f| \geq t}(x)t^p \leq \int_{X} |f|^p = \|f\|_{L^p(X, \mu)}^p$

and divide both sides by $t^p$.

This is a fundamental inequality in the the study of the interpolation of L^p spaces.

Answer by Damek Davis for More open problems

2011-03-29T17:39:06Z

Igor Shparlinski maintains a large list of open problems in exponential and character sums. You can find it on his website: http://web.science.mq.edu.au/~igor/

Is there a category structure one can place on measure spaces so that category-theoretic products exist?

2010-12-14T20:06:32Z

The usual category of measure spaces consists of objects $(X, \mathcal{B}_X, \mu_X)$, where $X$ is a space, $\mathcal{B}_X$ is a $\sigma$-algebra on $X$, and $\mu_X$ is a measure on $X$, and measure preserving morphisms $\phi \colon (X, \mathcal{B}_X, \mu_X) \to (Y, \mathcal{B}_Y, \mu_Y)$ such that $\phi_\ast \mu_X(E) = \mu_X(\phi^{-1}(E)) = \mu_Y(E)$ for all $E \in \mathcal{B}_Y$.

The category of measurable spaces consists of objects $(X, \mathcal{B}_X)$ and measurable morphisms $\phi \colon (X, \mathcal{B}_X) \to (Y, \mathcal{B}_Y)$.

Products exist in the category of measurable spaces. They coincide with the standard product $(X \times Y, \mathcal{B}_X \times \mathcal{B}_Y)$, where $X \times Y$ is the Cartesian product of $X$ and $Y$ and $\mathcal{B}_X \times \mathcal{B}_Y$ is the coarsest $\sigma$-algebra on $X\times Y$ such that the canonical projections $\pi_X \colon X \times Y \to X$ and $\pi_Y \colon X \times Y \to Y$ are measurable. Equivalently, $\mathcal{B}_X \times \mathcal{B}_Y$ is the $\sigma$-algebra generated by the sets $E \times F$ where $E \in \mathcal{B}_X$ and $F \in \mathcal{B}_Y$.

However, in the category of measure spaces, products do not exist. The first obstacle is that the canonical projection $\pi_X \colon X \times Y \to X$ may not be measure preserving. A simple example is the product of $(\mathbf{R}, \mathcal{B}[\mathbf{R}], \mu)$ with itself, where $\mathcal{B}[\mathbf{R}]$ is the Borel $\sigma$-algebra on $\mathbf{R}$. In this case,

$(\pi_{\mathbf{R}})_\ast\mu\times \mu([0,1]) = \mu\times \mu(\pi_\mathbf{R}^{-1}([0,1])) = \mu \times \mu([0,1]\times \mathbf{R}) = \infty \neq 1 = \mu([0,1])$ .

In addition, there may be multiple measures on $X\times Y$ whose pushforwards on $X$ and $Y$ are $\mu_X $ and $\mu_Y$. Terry Tao mentions that from the perspective of probability, this reflects that the distribution of random variables $X$ and $Y$ is not enough to determine the distribution of $(X, Y)$ because $X$ and $Y$ are not necessarily independent.

Given that the products in the usual category fail to exist, is it possible to define a new categorical structure on the class of measure spaces such that products do exist?

Comment by Damek Davis

2010-12-15T04:05:42Z

The counter example is that $\Delta^{-1}(\{(x,x) x\in [0,1]\}) = [0,1]$ has measure $1$ while the diagonal has measure 0.

Comment by Damek Davis

2010-12-15T04:05:32Z

No Problem :). I added a comment to clarify the meaning of $\mathcal{B}_X \times \mathcal{B}_Y$. If we cannot relax the condition of "measure preserving maps," then the product of the two spaces $X$ and $Y$ cannot be $X\times Y$ in general. Indeed, this forces the condition that $\pi_X$ and $\pi_Y$ are measurable. This forces the $\sigma$-algebra on $X\times Y$ to contain $\mathcal{B}_X \times \mathcal{B}_Y$. Thus, the counterexample of the diagonal map $\Delta : \mathbf{R} \to \mathbf{R} \times \mathbf{R}$ is available to us, a contradiction.

Comment by Damek Davis

2010-12-15T02:38:19Z

Even if we restrict to probability spaces and use the standard Carathéodory construction of product measure, the diagonal map $\Delta \colon [0,1] \to [0,1]\times [0,1]$ is not Borel measure preserving. Indeed, the line $E = \{(x,x) : x \in [0,1]\}$ has measure zero, while $\Delta^{-1}(E) = [0,1]$ has measure one.

Comment by Damek Davis

2010-12-15T01:01:50Z

Whoops! I meant to say that $\Delta^{-1}(\{y=x\}) = [0,1]$. This has measure $1$.

Comment by Damek Davis

2010-12-15T01:00:33Z

What about the line $\{y=x\}$ in $\mathbf{R}^2$ for $x \in [0, 1]$. This has Borel measure zero, but $\mu_R(\Delta^{-1}(\{y=1\})) = \mu_R([0,1]) = 1$