For $\phi(x,y)=(x,y+\psi (x))$ with $\psi:\Bbb{R}\to \Bbb{R}$ integrable, show that $\phi (B)$ is measurable for every box $B\subset \Bbb{R}^2$

Question

Let $\psi:\Bbb{R}\to \Bbb{R}$ be integrable and define $\phi:\Bbb{R}^2\to \Bbb{R}^2$ by $\phi(x,y)=(x,y+\psi (x))$. Prove that for every box $B\subset \Bbb{R}^2$, $\phi(B)$ is measureable and $v(\phi (B))=v(B)$.

Remarks:

• In the original question, the term used is "admissible" rather than "measurable"

• Admissibility of a set $A$ was conditioned, by definition, by the integrability of $1_A$.

I know that $\psi$ must be bounded, and I can picture $\phi(B)$ for $B=[a,b]\times [c,d]$, and it does seem intuitively like uncountably many shifting(for all $a\le x\le b$) in a way that preserves the volume of $B$. So, intuitively the volume of the second coordinate remains the same, but the first coordinate is an uncountable union, and I can't understand how to approach computing its volume. My ideal method would be using $1_A$ kind of function, but this time, $A$ is $x$-dependent. I also considered using $\int_{a}^{b}\int_{c+\psi(x)}^{d+\psi(x)}1dydx$, but I have no way to justify it and could use your help.

Answer 1

Here we go: I think you mean $\nu=\lambda^2$, the Lebesgue measure on $\mathbb{R}^2$.

$\nu(\phi(B)) = \int_{\mathbb{R}^2}\mathbf{1}_{\phi(B)}d\lambda^2=\int_{\mathbb{R}}\int_{\mathbb{R}}\mathbf{1}_{\phi(B)}d\lambda d\lambda=$

$=\int_{\mathbb{R}}\int_{\mathbb{R}}\mathbf{1}_{\phi(B)}(x,y)dy dx=\int_{\mathbb{R}}\int_{\mathbb{R}}\mathbf{1}_{\phi(B)}(x,y+\psi(x))dy dx=\int_{\mathbb{R}}\int_{\mathbb{R}}\mathbf{1}_{B}(x,y)dy dx=\int_{\mathbb{R}}\int_{\mathbb{R}}\mathbf{1}_{B}d\lambda d\lambda=\int_{\mathbb{R}^2}\mathbf{1}_{B}d\lambda^2=\nu(B)$ The first equality and the last are by definition, the second follows from Fubinis theorem, the third from the transformation formular, the rest are definitions again.

I assume there will be questions left, go ahead.

Answer 2

I see you already know that $\nu([a,b])=b-a$, so $\nu([a+\psi(x),b+\psi(x)])=b+\psi(x)-(a+\psi(x))=b-a$ for any fixed $x$.

Looks worse than it is:

$L_N(\mathbf{1}_{[a,b]\times[c+\psi(x),d+\psi(x)]}(x,y))=2^{-2N}\sum_{k,l \in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\inf_{y \in \frac{l}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]\times[c+\psi(x),d+\psi(x)]}(x,y)=2^{-2N}\sum_{k,l \in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\inf_{y \in \frac{l}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]\times[c,d]}(x,y+\psi(x))=2^{-2N}\sum_{k \in \mathbb{Z}}\sum_{l\in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\inf_{y \in \frac{l}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]}(x)\mathbf{1}_{[c,d]}(y+\psi(x)) =^{*} 2^{-2N}\sum_{k \in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\sum_{l\in \mathbb{Z}}\inf_{y \in \frac{l}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]}(x)\mathbf{1}_{[c,d]}(y+\psi(x)) = 2^{-N}\sum_{k \in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]}(x)2^{-N}\sum_{l\in \mathbb{Z}}\inf_{y \in \frac{l}{2^{N}}+[0,1]}\mathbf{1}_{[c,d]}(y+\psi(x)) =2^{-N}\sum_{k \in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]}(x)L_N(\mathbf{1}_{[c+\psi(x),d+\psi(x)]}(y))\geq2^{-N}\sum_{k \in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]}(x)L_N(\mathbf{1}_{(c+\psi(x),d+\psi(x))}(y))\geq 2^{-N}\sum_{k \in \mathbb{Z}}\inf_{x \in \frac{k}{2^{N}}+[0,1]}\mathbf{1}_{[a,b]}(x)((d+\psi(x)-(c+\psi(x))-2^{-N+1})=L_N(\mathbf{1}_{[a,b]}(x))((d-c)-2^{-N+1})\rightarrow (b-a)(d-c)$ as $ N \rightarrow \infty$ The last inequality is due to the hint on page 58. Use the same hint to show in the same fashion:

$U_N(\mathbf{1}_{[a,b]\times[c+\psi(x),d+\psi(x)]}(x,y)) \leq U_N(\mathbf{1}_{[a,b]}(x))(d-c+2^{-N+1}) \rightarrow (b-a)(d-c)$

The first inequality was just so you could refer to an open ball like you have in the hint, you wont need it for the $U_N$.

I have to work $=^{*}$ out, made a little mistake, give me some time...

The reason for $=^{*}$ is, that for any fixed $N$ looking at the indicator functions we see that only finitely many $k,l$ can result in a value other than 0, so there are some finite sets $A,B$ of natural numbers, such that $\sum_k \in \mathbb{Z}$ and $\sum_l \in \mathbb{Z} $ might be replaced with $\sum_k \in A$ and $\sum_l \in B $ then switching $\inf$ and finite sum is ok and then you just replace the sets back to $\mathbb{Z}$, because this means adding zero.