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$\textbf{Theorem:}$ Let $A,B$ two locally compact groups and $G=A\times B$ their product. Let $\mu_A$ and $\mu_B$ a left Haar measures. Then $\mu_A \times \mu_B$ is a left Haar measure of $G$.

$\textbf{Corollary:}$ Let $\bigtriangleup_A$ and $\bigtriangleup_B$ the modular functions of $A$ and $B$. Then

$$ \bigtriangleup_G(a,b)=\bigtriangleup_A(a)\bigtriangleup_B(b)$$

My proof:

Suppose that $A$ and $B$ are $\sigma-$finite then $\mu_A \times \mu_B$ satisfies:

$$ (\mu_A \times \mu_B) (M) = \int_A \mu_B (M_x) \mu_A(dx) $$

where $M$ is a borel set of $\mathcal{B}(A) \times \mathcal{B}(B)$ and $M_x = \{y \in B \hspace{0.5mm} \vert \hspace{0.5mm} (x,y) \in M \}$. It's easy to see that for $g=(g_1,g_2)\in G$ we have:

$$ (gM)_x = g_2 M_{g_1^{-1}x}$$

So:

$$ (\mu_A \times \mu_B) (gM) = \int_A \mu_B ((gM)_x) \mu_A(dx) = \int_A \mu_B (g_2 M_{g_1^{-1}x}) \mu_A(dx) = \int_A \mu_B (M_{g_1^{-1}x}) \mu_A(dx) $$

Then if put $y=g_1^{-1}x$ we have $\mu_A(dy)=\mu_A(dx)$ and:

$$ (\mu_A \times \mu_B )(gM) = \int_A \mu_B (M_{y}) \mu_A(dy) = (\mu_A \times \mu_B)(M)$$

The corollary is immediate.

My question is how to prove it for the case where it is not $\sigma$-finite. The theorem statement does not mention that $\mu_A$ and $\mu_B$ are $\sigma$-finite.

Bernard
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Leo
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1 Answers1

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This is not true as written.

For example, if $A$ is a locally compact Hausdorff group of cardinality greater than $\mathfrak c$, then $A\times A$ is Hausdorff, so the diagonal is Borel. On the other hand, it is not in the product sigma algebra $\mathcal B(A)\otimes \mathcal B(A)$ (see e.g. here).

It follows that $\mu_A\otimes \mu_A$ is not a Borel measure on $A\times A$, so it is not the Haar measure, even though it is left invariant on the product algebra (which follows easily from the definition), as well as inner and outer regular.

It is true if $A,B$ are second-countable (for example, they are real or complex Lie groups with countably many connected components), however, since in this case $\mathcal B(A)\otimes \mathcal B(B)=\mathcal B(A\times B)$. For locally compact groups, $\sigma$-finiteness follows from second countability (in fact, I think the two are equivalent, but I did not check this).

The Haar theorem implies that in general, $\mu_A\otimes \mu_B$ can be uniquely extended to a Haar measure on $A\times B$, and so the corollary is also true with no hypotheses about $A$ and $B$.

tomasz
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  • Any hint to prove that if $A$ and $B$ are second contable then: $\mathcal{B}(A) \times \mathcal{B}(B) = \mathcal{B}(A \times B)$. And why does Haar's theorem imply that there is an extension? – Leo Sep 05 '20 at 23:00
  • Because then you only need countable unions of open rectangles to obtain arbitrary open sets. To see the existence of an extension: choose any Haar measure $\mu'$ on $A\times B$. Choose an open rectangle $A_0\times B_0$ such that $\mu_A(A_0),\mu_B(B_0),\mu'(A_0\times B_0)<\infty$, and take $\mu=\frac{\mu_A(A_0)\mu_B(B_0)}{\mu'(A_0\times B_0)}\mu'$. Then $\mu$ is also a Haar measure and you can check that it extends $\mu_A\otimes \mu_B$. – tomasz Sep 05 '20 at 23:08
  • Thanks, one last question. If it is guaranteed that $\mathcal{B}(A) \times \mathcal{B}(B)=\mathcal{B}(A\times B)$ my proof is correct ?. Thanks again for giving you the time to answer me. – Leo Sep 05 '20 at 23:11
  • @Leo: It seems okay, but I think it is a bit overly complicated. Instead of applying Fubini, you can just directly argue that it is invariant by showing that it is true for rectangles, and then arguing that it follows that it is true for arbitrary Borel sets. Also, technically, you should also argue that the product measure is regular. – tomasz Sep 05 '20 at 23:24
  • Thank you very much. – Leo Sep 05 '20 at 23:28