Isomorphism of Measure Spaces

Motivation

I don't know the inception of isomorphic measure spaces, but it seems natural nowadays to look at the objects and morphisms of measurable spaces. The reason I become interested in this category was the following. We learned in topology that the standard topology on $\mathbb {R}$ is generated by a basis. This basis is the collection of all open intervals $(a,b)\{x\in \mathbb {R} :a<x<b,a,b\in \mathbb {R} \}$.

Definition

Let $X$ be a measurable space and $A$ a sigma algebra on $X$. Similary, Let $Y$ be a measurable space and $B$ a sigma algebra on $Y$. Let $(X,A)$ and $(Y,B)$ be measurable spaces.

• A map $f:X\rightarrow Y$ is called measurable if $f^{-1}(B)\in A$ for every $B\in B$.

• These two measurable spaces are called isomorphic if there exists a bijection $f:X\rightarrow Y$ such that $f$ and $f^{-1}$ are measurable (such $f$ is called an isomorphism).

Basic Theorem

Let $X_{1}$ and $X_{2}$ be Borel subsets of complete separable metric spaces. For the measurable spaces $(X_{1},B(X_{1}))$ and $(X_{2},B(X_{2}))$ to be isomoprhic, it is necessary and sufficient that the sets $X_{1}$ and $X_{2}$ be of the same cardinality.

Properties

We seek to find maps that preserve "essential" structures between measure spaces. Intuitively, we want at the minimum, maps to send sets of measure zero to sets of measure zero.

Smooth maps send sets of measure zero to sets of measure zero

Let $U$ be an open set of $\mathbb {R} ^{n}$, and let $f\colon U\rightarrow \mathbb {R} ^{n}$ be a smooth map. If $A\subset U$ is of measure zero, then $f(A)$ is of measure zero.

• Note that we can loosen the restriction of our map being smooth. It is enough to consider absolutely continuous functions and the statement still holds.

Mini-Sards Theorem

Let $U$ be an open set of $\mathbb {R} ^{n}$, and let $f\colon U\rightarrow \mathbb {R} ^{m}$ be a smooth map. Then if $m>n$, $f(U)$ has measure zero in $\mathbb {R} ^{m}$.

Example

Consider $f:\mathbb {R} \rightarrow \mathbb {R} ^{2}$ where $f(x)=(x,0)$. $f$ is easily seen to be a smooth map since it has partial derivatives of all order. Let $A=\{(x,0):x\in \mathbb {R} \}$. Pick $\epsilon >0$. Consider the cover $I_{k}=[k,k+1]\times [-\epsilon 2^{-|k|-4},\epsilon 2^{-|k|-4}]$. Then $A$ is covered by the union of all $J_{k}$. Now, $m(A)\leq \sum _{k\in \mathbb {Z} }m(J_{k})=\epsilon \sum _{k\in \mathbb {Z} }2^{-|k|-3}\leq \epsilon$. Since $\epsilon$ was aribtary $m(A)=0$.