Isomorphism of Measure Spaces

Personal 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 standard topology is connected but the lower limit topology is not connected. So now consider Theorem 1.18 from[1],

If $E\in M_{\mu }$, then $\mu (E)=\inf\{\mu (U):E\subset U{\text{ and }}U{\text{ is open }}\}=\sup\{\mu (K):K\subset E{\text{ and }}K{\text{ is compact }}\}$

This construction sort of has the same flavor as our topology example. But in this case, we get a positive result.

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).

Theorem

[If $X$ is a topological space, then $B(X)$ denotes the sigma algebra of Borel subsets of $X$.]

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.

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 $m(A)=0$.

Reference

1. Folland, Gerald. B; "Real Analysis: Modern Techniques and Their Applications." Wiley. 2007.

2. Victor Guillemin, Alan Pollack; "Differential Topology." Prentice-Hall. 1974.