Manifold Homotopies and Measures

Lee Smooth Manifolds

Some measure brainteasers (boo) followed by more manifolds.

(Folland 2.3) If $\{f_n\}$ is a sequence of measurable functions on $X$, then $F=\{x:\lim f_n(x)\text{ exists}\}$ is a measurable set.

Proof) We work in $\overline{\mathbb{R}}$ and then extend to $\mathbb{R}$ and $\mathbb{C}$. Let $g=\mathrm{limsup}{f}_n(x)$ and $h=\mathrm{liminf}{f}_n(x)$; $g$ and $h$ are clearly well-defined and measurable. Let $F^{\prime }=\{x:g(x)=h(x)=\pm \infty \}$, which is measurable. On $X\setminus F^{\prime }$, $g-h$ is well-defined and thus measurable, and $F=(g-h{)}^{-1}{|}_{X\setminus F^{\prime }}(\{0\})\cup F^{\prime }$ which is also measurable in $X$ as $F^{\prime }\in \mathcal{M}$. To extend this to $\mathbb{R}$, we can note that $G=\{x:f_n(x)\text{ is bounded}\}$ is measurable and $G\cap F$ is thus also measurable. $\mathbb{C}$ trivially follows by identifying $\mathbb{C}$ with ${\mathbb{R}}^2$.

(Folland 2.11) Suppose that $f$ is a function on $\mathbb{R}\times {\mathbb{R}}^k$ such that $f(x,\cdot )$ is Borel measurable for each $x\in \mathbb{R}$ and $f(\cdot ,y)$ is continuous for each $y\in {\mathbb{R}}^k$. For $n\in \mathbb{N}$, define $f_n$ as follows. For $i\in \mathbb{Z}$ let $a_i=i/n$, and for $a_i\le x\le a_{i+1}$ let

$$f_n(x,y)=\frac{f(a_{i+1},y)(x-a_i)-f(a_i,y)(x-a_{i+1})}{a_{i+1}-a_i}.$$

Then $f_n$ is Borel measurable on $\mathbb{R}\times {\mathbb{R}}^k$ and $f_n\to f$ pointwise; hence $f$ is Borel measurable on $\mathbb{R}\times {\mathbb{R}}^k$. Conclude by induction that every function on ${\mathbb{R}}^n$ that is continuous in each variable separately is Borel measurable.

Proof) It is trivial that $f_n\to f$ pointwise. In each interval $x\in [a_i,a_{i+1}]$, $f_n(x,y)$ is an algebraic combination of Borel measurable sets and thus is itself Borel measurable. It suffices to show that, if a sequence of Borel measurable functions $f_n$ on ${\mathbb{R}}^n$ converges pointwise to $f$, then $f$ must be Borel measurable. With range $\overline{\mathbb{R}}$ this is a well-known fact; with range $\mathbb{R}$ it also works as each sequence $f_n(x)$ is bounded and thus has well-defined $\mathrm{limsup}$ and $\mathrm{liminf}$. $\mathbb{C}$, as usual, follows from $\mathbb{R}$. The final line is textbook induction.

(Lee 9.26) Let $M$ be a smooth manifold with nonempty boundary, and let $\iota :\text{Int }M\to M$ denote inclusion. There exists a proper smooth embedding $R:M\to \text{Int }M$ such that both $\iota \circ R$ and $R\circ \iota$are smoothly homotopic to identity maps. Therefore, $\iota$ is a homotopy equivalence.

Proof) $\partial M$ has a collar neighborhood $C$ in $M$ that, as a submanifold, is diffeomorphic to $[0,1)\times \partial M$ via a smooth embedding $E:[0,1)\times \partial M\to M$. The problem with doing topology with $C$ is that $E([0,a]\times \partial M)$ is closed in $C$, but this doesn’t imply closedness in $M$. We thus have to refine $C$ to be better-behaved; let $f:M\to {\mathbb{R}}^{+}$ be a positive smooth exhaustion function and $W:\{(t,x):f(E(t,x))>f(x)-1\}$. $W$ is a neighborhood of $C$, and using partitions of unity we can construct $\delta :\partial M\to {\mathbb{R}}^{+}$ such that $\delta <1$ and $[0,\delta )\times \partial M\subset W$. Defining $\hat{E}:[0,1)\times \partial M\to M$ such that $\hat{E}(t,x)=(t\delta (x),x)$ and $\hat{C}=\hat{E}([0,1)\times \partial M)$ we obtain another collar neighborhood $\hat{C}$ that is suitable for our purposes, as $\hat{E}([0,a]\times \partial M)$ is closed in $M$ for all $0<a<1$. We prove this as follows; let $p$ be a limit point of $\hat{E}([0,a]\times \partial M)$ and let $\hat{E}(t_n,x_n)\to p$ where $(t_n,x_n)\in [0,a]\times \partial M$. $f(\hat{E}(t_n,x_n))$ is then bounded, which means $f(x_n)$ is bounded. Note that $\partial M$is closed in $M$, and thus $f{|}_{\partial M}$ is a smooth positive exhaustion of $\partial M$; thus $(t_n,x_n)$ has a convergent subsequence that converges to $(t,x)\in [0,a]\times \partial M$, forcing $p=\hat{E}(t,x)\in \hat{E}([0,a]\times \partial M)$. We are done; from now on we replace $C$ and $E$ with $\hat{C}$ and $\hat{E}$. We also identify $\hat{E}(t,x)\in \hat{C}$ simply with $(t,x)$ for sanity.

Let $C(a)=\{(t,x)\in C|0\le t<a\}$ and $M(a)=M\setminus C(a)$. Then, $M(a)$ must be a regular domain of $M$ for all $0<a<1$; if $p\in M(a)$ is such that $p\notin C(a,\partial M)$, $p$ is in the topological interior of $M(a)$ (by the argument above), and if $p=C(a,x)$ for some $x\in \partial M$, $C$ is a neighborhood of $p$ such that $C\cap M(a)$ is the diffeomorphic image of $[a,1)\times \partial M$. By the slice chart lemma for submanifolds with boundary, $M(a)$ is a regular domain.

Let $\psi :[0,1)\to [1/3,1)$ be an increasing diffeomorphism that is the identity on $[2/3,1)$. Define $R:M\to \text{Int }M$ by $R(p)=p$ if $p\in \text{Int }M(2/3)$ and $(\psi (t),x)$ if $p=(t,x)\in C$. This is a diffeomorphism onto the regular domain $M(1/3)$ of $\text{Int }M$, and is the desired proper smooth embedding. Defining $H:M\times I\to M$ by $H(p,t)=p$ if $p\in \text{Int }M(2/3)$ and $H(p,t)=(ts+(1-s)\psi (t),x)$ if $p=(s,x)\in C$, we have the desired smooth homotopies for $\iota \circ R$ and $R\circ \iota$.