Timeline for Are measures better thought of as densities than differentials?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 1, 2020 at 12:36 | comment | added | Michael Greinecker | The Radon-Nikodym derivative has its name obviously from the fundamental theorem of calculus. I have never seen someone getting confused by the name. | |
| Jun 1, 2020 at 11:48 | comment | added | Dieter Kadelka | For me its a strange discussion. The proposal seems only to make sense if we have probability measures on $\mathbb{R}$ or some locally compact group. But what if we have a measure on an abstract measure space $(\Omega,\cal{A})$? Then $dx$ with respect to what? | |
| Jun 1, 2020 at 11:34 | review | Close votes | |||
| Jun 7, 2020 at 19:26 | |||||
| Jun 1, 2020 at 8:53 | comment | added | leo monsaingeon | Well, this is some kind of residual notational ambiguity due to the fact that, when one learns integration, usually the Lebesgue integral is given a priviledged role (for obvious historical and convenience reasons). For example (some) probabilists would never actually write $\int f(x) dx$, but rather $\int f(x)\lambda(dx)$ or $\int f(x) \mathcal L^d(dx)$, where $\lambda$ or $\mathcal L^d$ stand for the $d$-dimensional Lebesgue measure. Actually the second notation is somehow also standard in geometric measure theory and calculus of variations, especially in the Italian school. | |
| Jun 1, 2020 at 8:50 | comment | added | wlad | @leomonsaingeon I guess that makes sense. But now you can't write $\int f(x) \, dx$ because $dx$ is some "infinitesimal" set. You have to write $\int f(x) \mu(dx)$. But I guess that apart from that awkwardness, it could work | |
| Jun 1, 2020 at 8:47 | comment | added | leo monsaingeon | yes, precisely! the only difference between a "sum" and the more rigorous integral formalism is that $\mu(dx)$ is the ininitesimal mass contained in the infinitesimal set $dx$ (in fact as far as I can tell probabilists often write $dx$ for measurable sets, so non infinitesimal at all) After all, an integral is nothing but a fancier weighted sum, is it not? (the $\int$ sign is even reminiscent from that idea) | |
| Jun 1, 2020 at 8:44 | comment | added | wlad | @leomonsaingeon To me that reads "Sum of $f(x)$ times $\mu(dx)$" where $\mu(dx)$ doesn't make much sense on its own, but maybe I'm being too literal. I like to think of $\int f(x)\,dx$ as literally being the sum of $f(x)$ times $dx$ | |
| Jun 1, 2020 at 8:42 | comment | added | leo monsaingeon | What would be the advantage, really? Although I'm an analyst I personally also like the probabilistic notation $\int f(x)\mu(dx)$ | |
| Jun 1, 2020 at 8:04 | history | asked | wlad | CC BY-SA 4.0 |