Isomorphism of Measure Spaces

Motivation

The reason I become interested in this category was the following. We learned that the standard topology on $\mathbb {R}$ is generated by a basis. This basis is the collection of all open intervals $(a,b):=\{x\in \mathbb {R} :a<x<b,a,b\in \mathbb {R} \}$. Now, consider the collection A of subsets in $\mathbb {R}$: $A:=\{[a,b):=\{x\in \mathbb {R} |a\leq x<b\}a,b\in \mathbb {R} \}$. The topology generated by the half-open intervals, called the lower limit topology, yields a different topology on the real numbers than the standard topology. We can see this by recalling that the real numbers under the standard topology is connected but the lower limit topology is not connected.

Definition

Let $X$ be a measurable space and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} a sigma algebra on $X$. Similary, Let $Y$ be a measurable space and $B$ a sigma algebra on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y} . Let $(X,A)$ and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m > n } , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(U)} has measure zero in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}^m} .

Example

Consider $f:\mathbb {R} \rightarrow \mathbb {R} ^{2}$ where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=(x,0)} . $f$ is easily seen to be a smooth map since it has partial derivatives of all order. Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 $m(A)=0$.