Show that $\lambda^d$ is invariant under skew transformation.

Question

In my book we have shown that the Lebesgue-Borel measure $\lambda^d$ is the only translation invariant measure on $\mathcal{B}^d$ with $\lambda^d(W)=1$, where $W=[\boldsymbol{0},\boldsymbol{1})$. Also, if $D^{(k)}_{\alpha}:\mathbb{R}^d \to \mathbb{R}^d$ is the map which multiplies the $k$th coordinate of a point in $\mathbb{R}^d$ by the nonzero scalar $\alpha$, then we have shown that $\lambda^d((D^{(k) }_{\alpha})^{-1}(A))=|\alpha|^{-1}\lambda^d(A)$ for all $A\in \mathcal{B}^d$.

Given $i,k\in \{ 1,\dots,d\}$ with $i\neq k$, we now consider the skew map $S:\mathbb{R}^d \to \mathbb{R}^d$ defined by

$$S(x_1,\dots,x_d)=(x_1,\dots,x_{i-1},x_i+x_k,x_{i+1},\dots,x_d)$$

and we want to show that $$\lambda^d(S^{-1}(A))=\lambda^d(A)$$ for all $A\in \mathcal{B}^d$. We note that $S$ is continuous, hence measurable, and invertible, with inverse $S^{-1}=D^{(k) }_{-1}\circ S \circ D^{(k) }_{-1}$.

My book now makes the following two claims:

$1.$ Since $S$ is a linear mapping, $\lambda^d(S^{-1}(\cdot))$ is a translation invariant measure on $\mathcal{B}^d$, and so the result will follow if we can show that $\lambda^d(S^{-1}(W))=1$.

$2.$ Given the equality $S^{-1}=D^{(k) }_{-1}\circ S \circ D^{(k) }_{-1}$, it suffices to show instead that $\lambda^d (S(W))=1$.

I believe I understand the first claim: since $S$ and $D^{(k) }_{-1}$ are both linear, we have that $S^{-1}$ is linear, and so the translation invariance of $\lambda^d(S^{-1}(\cdot))$ follows from the translation invariance of $\lambda^d$. The result then follows from $\lambda^d(S^{-1}(W))=1$ by the uniqueness result first mentioned.

But I don't undertsand why the second claim is true. Any help would be greatly appreciated.

Answer 1

I'll declutter the notation, and ask why when $S^{-1}=D\circ S\circ D$ then $\lambda(S(W))=1$ entails $\lambda(S^{-1}(W))=1$. (I think that's the problem).

Suppose $\lambda(S(W))=1$. Then $D$ preserves measure (you've shown this). If $\lambda(S(W))=1$ then $S$ preserves measure, as $X\mapsto\lambda(S(X))$ is a translation-invariant measure. So $D\circ S\circ D$ preserves measure: $S^{-1}$ preserves measure.

But it's obvious that $S$ preserves measure iff $S^{-1}$ does (isn't) it. And it's no easier to prove $S^{-1}(W)$ has measure $1$ than it is to prove $S(W)$ has, so I don't see why this book is doing it in this rather peculiar manner.

Perhaps I'm missing something.

Answer 2

Disclaimer: This is not exactly what was asked and it is a prove that requires knowing Lebesgue integral and Tonelli's Theorem, but makes the statement quite clear.

It can also be proven using Tonelli's Theorem for measurable, $f\geq 0$ and thus for every Borel measurable $A\subset \mathbb{R}^d$. Of course it is used that $\lambda^2 = \lambda \otimes \lambda$.

I'll do it for the case $d=2$ and for $S_1: \mathbb{R}^2\to \mathbb{R}^2$ where $S_1(x,y)=(x+y, y)$, so it's clear of what's going on. The same arguments can be applied to prove the general case. Let $f\geq 0$, measurable. Then, $f\circ S_1 \geq 0$ and measurable and, \begin{align} \int_{\mathbb{R}^2} f\circ S_1(x,y) \text{ d}\lambda^2 = \int_{\mathbb{R}^2} f(x+y,y) \text{ d}\lambda^2 &\overset{\text{Tonelli}}{=} \int_{\mathbb{R}} \int_{\mathbb{R}} f(x+y,y) \text{ d}\lambda(x) \text{ d}\lambda(y) \\&\overset{(1)}{=} \int_{\mathbb{R}} \int_{\mathbb{R}} f(x,y) \text{ d}\lambda(x) \text{ d}\lambda(y) \overset{\text{Tonelli}}{=} \int_{\mathbb{R}^2} f\text{ d}\lambda^2 \end{align}

Equality $(1)$ holds since, $\lambda$ (Lebesgue measure on the line) is invariant under translations.