## Tuesday, January 08, 2013

### MathJaxing!

This is a test post to check if MathJax rendering of Latex has been effectively integrated into Blogger.

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The following is a well known point about completion of product measures:

Consider measure spaces $(X,\Sigma_X,\mu)$ and $(Y,\Sigma_Y,\mu)$ where $\mu$ is the Lebesgue measure. For the sake of definiteness, let's set the spaces $X$ and $Y$ as the unit interval $[0,1]$ and the sigma algebras $\Sigma_X$ and $\Sigma_Y$ as the standard sigma algebra of Lebesgue measurable sets on $[0,1]$.

Now we want extend these "coordinate" measures to the product space $X \times Y$. A natural way to do so is to define a product sigma algebra on $X \times Y$ by forming the smallest sigma algebra containing all sets of the form $A \times B$, where $A\in\Sigma_X$ and $B\in\Sigma_Y$ are measurable sets on the $x$ and $y$ coordinate spaces respectively. (For our example, these are just the intervals $X=[0,1]$ and $Y=[0,1]$.)

That is, the product space on which we are defining a product measure is:

$$(X\times Y;\sigma\{A\times B:A\in \Sigma_X, B\in \Sigma_Y\}; \tilde{\mu})$$

For a measurable rectangle $A\times B$,

$$\tilde{\mu}(A\times B)=\mu(A)\mu(B)$$

In the spirit of this product measure space, let's see what happens once we put $A=\{x_0\}$ (just one point). Then of course, our standard construction yields:

$$\tilde{\mu}(A\times B)=\mu(A)\mu(B)=0\cdot \mu(B)$$

for any set $B\in 2^{Y}$, not just in the sigma algebra $\Sigma_Y$.

And here is where all the trouble starts! How so? Well for all sets in the sigma algebra $\Sigma_Y$,

$$\tilde{\mu}(A\times B)=\mu(A)\mu(B)=0\cdot \mu(B)=0$$

but for all sets $B\notin \Sigma_Y$ (and their existence is guaranteed - Vitali sets are an example), the above equation is not well defined! However the set $\{x_0\}\times B\subseteq \{x_0\}\times [0,1]$ and this larger set does have measure 0!

Hence the product measure $\tilde{\mu}$ must be completed as well! We can do this in the standard way: take all null sets with respect to the new measure $\tilde{\mu}$ and construct a new sigma algebra generated by such sets and the old product sigma algebra.

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Yes, it has!