Question

Why on a complete metric spaces that are Polish spaces we can define probability measures?

score 210 · Answer 1 · 2019-07-19 09:09:53

210

Views

Why on a complete metric spaces that are Polish spaces we can define probability measures?

Published on 19 Jul 2019 - 9:09

score 41 · Answer 2 · 2019-08-23 11:02:20

41

Views

Proposition 8.3.4 in Donald L. Cohn's Book "Measure Theory" , page 258

Published on 23 Aug 2019 - 11:02

score 275 · Answer 3 · 2019-09-07 11:49:50

275

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$X$ Polish space. Homeomorphism between $X$ and subset of $[0,1]^{\mathbb{N}}$

Published on 07 Sep 2019 - 11:49

score 205 · Answer 4 · 2019-09-17 13:18:17

205

Views

Tight subsubsequence implies tight

Published on 17 Sep 2019 - 13:18

score 252 · Answer 5 · 2019-09-20 23:30:38

252

Views

Existence of increasing sequence of compact sets

Published on 20 Sep 2019 - 23:30

score 161 · Answer 6 · 2019-09-22 21:35:25

161

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Let $X \subset \mathbb{R}$ be $G_{\delta}$ is homeomorphic to $\mathcal{N}$.

Published on 22 Sep 2019 - 21:35

score 71 · Answer 7 · 2019-10-01 18:52:58

71

Views

Proving a metric space is Polish. Finished and looking for feedback

Published on 01 Oct 2019 - 18:52

score 149 · Answer 8 · 2019-10-01 20:30:51

149

Views

Space $M(\lambda)$ of measurable subsets of $[0,1]$ up to null sets is Polish

Published on 01 Oct 2019 - 20:30

score 692 · Answer 9 · 2019-10-02 08:47:29

692

Views

Definition of Suslin space

Published on 02 Oct 2019 - 8:47

score 128 · Answer 10 · 2019-10-02 09:54:03

128

Views

Is the set of probability measures with strong convergence compact and/or metrizable?

Published on 02 Oct 2019 - 9:54

score 218 · Answer 11 · 2019-10-10 09:16:21

218

Views

Separable metrizable space as dense subset of a compact metrizable space

Published on 10 Oct 2019 - 9:16

score 490 · Answer 12 · 2019-10-31 15:18:27

490

Views

Is a polish space sigma compact?

Published on 31 Oct 2019 - 15:18

score 363 · Answer 13 · 2019-10-31 17:21:38

363

Views

Class of absolutely continuous probability measures is $G_\delta$

Published on 31 Oct 2019 - 17:21

score 164 · Answer 14 · 2019-11-19 23:05:06

164

Views

Why is $2^{\mathbb{N}}$ Homeomorphic to $2^{\mathbb{N}^{< \mathbb{N}}}$?

Published on 19 Nov 2019 - 23:05

score 56 · Answer 15 · 2019-11-21 08:01:11

56

Views

Weak compactness in measures implies compactness in the underlying metric space via the Dirac's delta

Published on 21 Nov 2019 - 8:01

List Question

Why on a complete metric spaces that are Polish spaces we can define probability measures?

Proposition 8.3.4 in Donald L. Cohn's Book "Measure Theory" , page 258

$X$ Polish space. Homeomorphism between $X$ and subset of $[0,1]^{\mathbb{N}}$

Tight subsubsequence implies tight

Existence of increasing sequence of compact sets

Let $X \subset \mathbb{R}$ be $G_{\delta}$ is homeomorphic to $\mathcal{N}$.

Proving a metric space is Polish. Finished and looking for feedback

Space $M(\lambda)$ of measurable subsets of $[0,1]$ up to null sets is Polish

Definition of Suslin space

Is the set of probability measures with strong convergence compact and/or metrizable?

Separable metrizable space as dense subset of a compact metrizable space

Is a polish space sigma compact?

Class of absolutely continuous probability measures is $G_\delta$

Why is $2^{\mathbb{N}}$ Homeomorphic to $2^{\mathbb{N}^{< \mathbb{N}}}$?

Weak compactness in measures implies compactness in the underlying metric space via the Dirac's delta

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