Isomorphism of Measure Spaces

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 isomoprhuic, it is necessary and sufficient that the sets $X_{1}$ and $X_{2}$ be of the same cardinality.

Properties

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 $m(A)=0$.