Timeline for Question on the Quotient Integral Formula
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 16, 2019 at 8:10 | comment | added | user130903 | No. Read the top of page 20 (new edition). | |
| Oct 15, 2019 at 20:12 | comment | added | Fabian Werner | I think (not 100% sure again) that he proves the quotient formula for bounded functions first (and in this special case he probably needs the bounded property) and then extends it in the way above... | |
| Oct 15, 2019 at 4:55 | comment | added | user130903 | Why should the function f be bounded? This is not necessary. Even unbounded functions are monotonous limits of bounded Lebesgue step functions. So, when you know the quotient formula for Lebesgue step functions, you know it for all functions in L1. | |
| Oct 13, 2019 at 21:35 | comment | added | Fabian Werner | I think I meant $I_f = I_g$ at least $X/G$-almost everywhere... | |
| Oct 13, 2019 at 21:23 | comment | added | Fabian Werner | $f(\mathcal{C}g)$, right? The function $\tilde{f}: X/G \mapsto \mathbb{R}, \tilde{f}(xG) := f(x)$ is not well defined. Hence we actually define $I_f(xG) := \int_G f(xg) dg$ which then is well defined. However, this expression needs to be independent of the chosen representative of the $L^1$ class, i.e. given $f=g$ $X$-almost everywhere, is $I_f = I_g$? And I cannot remember why right now but it must be a relationship between the measure on $X$ and the one on $X/G$ that makes this true... that was what I did not understand at the time... | |
| Oct 13, 2019 at 21:21 | comment | added | Fabian Werner | Depending on your teacher, $L^1$ functions might or might not take the value $\pm \infty$ itself. So let's assume that they are allowed to do that. Since they are in $L^1$ we know that the set of $x$ such that $f(x)=\infty$ is a zero set, however, without changing the function on that set you will never achieve that they are bounded. Not 100% sure anymore (It's been a few years :-)) but I think I did not feel comfortable with the relationships of the measures on $X$ and the one on $X/G$: What does the formula say? Well first of all we need to make sense of the expression | |
| Oct 13, 2019 at 2:28 | comment | added | user130903 | Why does one have to change $f$ in this argument? The $f_n$ are surely bounded, so once proven the quotient formula for them, it follows for $f$ by monotonous convergence, or does it? | |
| Jan 6, 2014 at 15:27 | history | asked | Fabian Werner | CC BY-SA 3.0 |