Timeline for the "observable" space of a measure space
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2017 at 0:55 | history | edited | burtonpeterj | CC BY-SA 3.0 |
Added the assumption that $X$ is standard.
|
| Nov 20, 2017 at 0:54 | comment | added | burtonpeterj | I should have assumed that the probability spaces are standard. | |
| Nov 14, 2017 at 15:40 | comment | added | Michael Greinecker | It is not true that "any two nonatomic probability spaces are isomorphic", not even in terms of their measure algebra. | |
| Nov 14, 2017 at 12:46 | comment | added | Nik Weaver | Actually, that isn't right. It only shows that the limit of a sequence of measurable functions is measurable. The limit of a net won't be, in general. And anyway the metric topology is different from the product topology. | |
| Nov 14, 2017 at 6:39 | comment | added | Shakiba | @nik-weaver oops! my reasoning was incorrect. Let me try a different reasoning: measurability of $f$ means $f^{-1} (a)$ is measurable for any $a \in F$ but when a sequence of measurable functions $f_{n}$ converges to $f$, $f^{-1}(a)=\{x \in X | \exists N, \forall n \geq N, x \in f_{n}^{-1}(a)\}$. Latest equality is nothing but $\bigcup_{k=1} \bigcap_{n=k}f_{n}^{-1}(a)$ which is measurable. So this is a proof for: pointwise limit of measurable observable is a measurable one. | |
| Nov 13, 2017 at 18:47 | comment | added | Shakiba | @nik-weaver For finiteness of $F$, any convergent sequence of measurable, $f_{n}$, which is convergent to a function $f$, when is computed in a point must be ultimately constant so there is a &n& such that $f_{n}=f$, thus $f$ is measurable. Am I wrong? | |
| Nov 13, 2017 at 18:15 | comment | added | Nik Weaver | @Shakiba, burtonpeterj is right. (The set of measurable elements is typically not closed.) | |
| Nov 13, 2017 at 17:49 | comment | added | Shakiba | For any set $X$ and finite set $F$, the topological space $F^{X}$ with product topology which is pointwise convergence, is a compact space (Tychonoff's theorem). But measurable elements if this space is a closed subset so is compact. | |
| Nov 13, 2017 at 15:07 | history | answered | burtonpeterj | CC BY-SA 3.0 |