*[Corrected, thanks – T.]*

It seems to me that the Lebesgue measure is invariant under addition, but not under the multiplication in the group. Don’t you need to divide the Lebesgue measure by the $d$-th power of the absolute value of the determinant to make it a Haar measure?

*[Fair enough; I have corrected the text accordingly -T.]*

Sorry to bother with such questions, I’m new to these topics.

Thanks in advance for your help!

*[The phrase “total mass” before Exercise 24 has been corrected to “total variation”. -T.]*

Yes, the first inequality in Exercise 13 is in fact an equality (and I have now adjusted the exercise accordingly), though the second inequality is not.

For Exercise 21(i), the averaging trick only works when G is compact, or at least amenable. Otherwise, the Haar measure is infinite and there is no obvious way to perform the averaging. (But this is certainly “morally” the right way to think about the problem.)

]]>Also, in Exercise 21(i), can’t we just take the Haar integral of \rho(g)v + c(g)? By convexity, the result must lie in the ball (and hence be finite), and by the left-invariance of the Haar measure, it has the behavior that we’re looking for.

]]>You’re right, that fixes it.

Still working my way through Proposition 9: Unless I’m not looking at it right, the end of the proof seems a little oversimplified. Namely, you claim that the -invariance of implies the -invariance of (for almost all ). That this is not the case can be seen from the case of finite: -invariance of only implies -invariance of (unless is a normal subgroup).

Actually -invariance of means that for all the equation holds for almost all . Now we must apply Fubini in order to change the order of quantification over and , to deduce that for almost all the equation holds for almost all , which then implies that is almost everywhere constant. At least, this seems to be the approach in the Bekka, de la Harpe, and Valette book, Theorem E.3.1.

*[Corrected, thanks – T.]*

Yes, this was a bit awkward; for my applications I only needed the compactly generated case and so assumed it throughout for simplicity, but for other applications (such as Kazhdan’s original application) it is important to not assume compact generation *a priori*. But for the specific task of proving Proposition 9, the compact generation of G is not actually needed; if one inspects the argument, one sees that the set S does not need to generate all of G, but merely to generate a certain compact subset of G, namely those h such that gh lies in the support of f for some g in K’. Indeed, one could basically just set S to be that set. I’ll adjust Remark 6 to reflect this.