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$\newcommand{\d}{\mathrm{d}}$ This the Exercise 3, Chapter 11 of the Gerald B. Folland book Real Analysis:

Let $G$ be a locally compact group that is homeomorphic to an open subset $U$ of $\Bbb R^d$ in such a way that, if we identify $G$ with $U$, left translation is an affine map -- that is, $xy=A_x(y)+b_x$ where $A_x$ is a linear transformation of $\Bbb R^d$ and $b_x\in\Bbb R^d$. Then $|\det A_x|^{-1}\d x$ is a left Haar measure on $G$, where $\d x$ denotes Lebesgue measure on $\Bbb R^d$.

Let me know if I understand this.

What the problem gives us is $G$ a locally compact group, an open set $U\subseteq\Bbb R^d$, and a bijection $\varphi:G\to U$ such that $\varphi$ and $\varphi^{-1}$ are both continuous and satisfy that given $u\in G$ there is a linear operator $A_{\varphi(u)}:\Bbb R^d\to \Bbb R^d$ and $b_{\varphi(u)}\in\Bbb R^d$ so that $$\varphi(uv)=A_{\varphi(u)}(\varphi(v))+b_{\varphi(u)}.$$

Is this right

If it is, define $f:\Bbb R^d\to\Bbb [0,\infty[$ given by $$f(x)=|\det A_x|^{-1}\quad\text{i.e.}\quad f(x)=|\det A_{\varphi(\varphi^{-1}(x))}|^{-1}.$$

Is the problem asking if the measure $\mu$ in $G$ given by $$\mu(E)=\int_{\varphi(E)}f(x)\d x$$ is a left Haar measure?

This reminds me the formula $$\int_E f(y)\d y=|\det T|\int_{T^{-1} E} f(Tx)\d x,$$ but I don't know what can I do.

Any advice in how to interpret the problem or on how to proceed is very appreciated.

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As to why such a $\phi$ exists: That's an assumption, part of the "Let ..."; no reason is given or required. – joriki Jul 18 '12 at 5:49
@joriki okay ${}{}$ – leo Jul 18 '12 at 5:56
What do you have to check for $\mu$ to be a Haar measure? ... – martini Jul 18 '12 at 6:15
@martini $\mu$ is a Haar measure if is a nonzero Radon measure that satisfies that for all $x\in G$ and $E\subseteq G$, $\mu(xE)=\mu(E)$ – leo Jul 18 '12 at 6:27
What of these condition did you manage to check? – martini Jul 18 '12 at 6:42
up vote 2 down vote accepted

The notation $xy=A_xy+b_x$ is quite confusing indeed. It appears on the left hand side, that $x,y\in G$; however, on the right hand side, it would appear that $y\in\mathbb{R}^d$. Your introduction of the homeomorphism, $\varphi:G\mapsto U\subset\mathbb{R}^d$, is the right thing to do.

Is this right

Yes, that is a restatement of part of the hypothesis with the homeomorphism given explicitly by $\varphi$. Since $\varphi$ is a homeomorphism, we could simply write $$ \varphi(uv)=A_u\varphi(v)+b_u $$

Is the problem asking if the measure $\mu$ in $G$ given by $$\mu(E)=\int_{\varphi(E)}f(x)\d x$$ is a left Haar measure?

Since $f(x)$ is meaningless for $x\not\in U$, I would define $f:U\mapsto[0,\infty)$ by $$ f(\varphi(u))=|\det A_u|^{-1} $$ and then define $\mu$ as you do above: $$ \begin{align} \mu(E) &=\int_{\varphi(E)}f(x)\,\mathrm{d}x \end{align} $$ Now note that $$ \begin{align} \varphi(uvw) &=\color{#C00000}{A_{uv}}\varphi(w)+\color{#00A000}{b_{uv}}\\ &=A_u\varphi(vw)+b_u\\ &=A_u(A_v\varphi(w)+b_v)+b_u\\ &=\color{#C00000}{A_uA_v}\varphi(w)+\color{#00A000}{A_ub_v+b_u} \end{align} $$ Therefore, we get $A_{uv}=A_uA_v$. Thus, we get $$ \begin{align} f(\varphi(uv)) &=|\det(A_{uv})|^{-1}\\ &=|\det(A_uA_v)|^{-1}\\ &=|\det(A_u)\det(A_v)|^{-1}\\ &=f(\varphi(u))f(\varphi(v)) \end{align} $$ At this point, to show the translation invariance, $$ \begin{align} \mu(uE) &=\int_{\varphi(uE)}f(x)\,\mathrm{d}x\\ &=\int_{\varphi(E)}\color{#C00000}{f(\varphi(u\varphi^{-1}(x)))}\,\color{#00A000}{\mathrm{d}(A_{u}x+b_{u^{-1}})}\\ &=\int_{\varphi(E)}\color{#C00000}{f(\varphi(u))f(x)}\color{#00A000}{|\det A_{u}|\,\mathrm{d}x}\\ &=\int_{\varphi(E)}|\det A_{u}|^{-1}f(x)|\det A_{u}|\,\mathrm{d}x\\ &=\int_{\varphi(E)}f(x)\,\mathrm{d}x\\ &=\mu(E) \end{align} $$

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This is great! thanks for your answer. – leo Jul 19 '12 at 1:39

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