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Jun
19
comment Subset of $\mathbb{I}\cap [0,1]$ (irrationals in [0,1]) that is closed in $\mathbb{R}$ and has measure $\epsilon \in (0,1)$
In what text did you see this and what page?
Jun
19
comment A simple question about open set
@Mathematics $B(-1,r)=\{x\in S:d(x,-1)<r\}$ as a nhood of $-1$ in $S$ is by definition a subset of $S$. Ofcourse the open ball in $X$ is not a subset of $S$.
Jun
18
comment Continuous images of open sets are Borel?
Thanks Brian, this answer is also very enlightening.
Jun
18
awarded  Nice Question
Jun
18
comment Continuous images of open sets are Borel?
@MichaelGreinecker: Thanks!
Jun
18
comment Continuous images of open sets are Borel?
@t.b. True, that would do the job. Thanks.
Jun
18
accepted Continuous images of open sets are Borel?
Jun
18
asked Continuous images of open sets are Borel?
Jun
17
comment Compactness in $\mathbb{Q}$
@MarcoCastronovo: You can do it the following way. Let $A\subset X$ and give $A$ the subspace topology of $X$. Show that $A$ is compact if and only if every cover of $A$ in $X$ has a finite subcover. The role of $X$ was arbitrary, as long as the topology of $A$ is the subspace topology from $X$.
Jun
17
comment Sigma-field of a sequence of Random Variables
Okey. Just as I had assumed too.
Jun
17
comment Sigma-field of a sequence of Random Variables
By the way, it is different from saying that we have a random variable $\tilde{X}:\Omega\to\mathbb{R}^{\infty}$ given by $\tilde{X}(\omega)=(X_{1}(\omega),X_{2}(\omega),...)$ than saying that $\tilde{X}=(X_{i})_{i=1}^{\infty}$ is a sequence of random variables. When looking at the $\sigma$-algebras they generate, in first case you need to consider preimages of Borel sets in $\mathbb{R}^{\infty}$ and in latter you need to consider preimages or Borel sets in $\mathbb{R}$ for each $X_{n}$. And whether these two agree depends on the topology of $\mathbb{R}^{\infty}$. Which one did you mean?
Jun
17
revised Sigma-field of a sequence of Random Variables
added 120 characters in body
Jun
17
answered Sigma-field of a sequence of Random Variables
Jun
16
comment Form of $\sigma(X_n)$ and $\sigma(X_n)$-measurability
@Justin: What I meant was, that $\sigma(\tilde{X})=\sigma(\{\{X_{n}\leq a\}:a\in\mathbb{R},\,\,n\in\mathbb{N}\})$, i.e. that $\{X_{n}\leq a\}$ form of sets are generators of $\sigma(\tilde{X})$. This is straight from the definition, as $\sigma(\tilde{X})$ is the smallest $\sigma$-algebra for which respect all the functions $X_{n}$ are measurable, and $X_{n}$ to be measurable is equivalent of saying that $\{X_{n}\leq a\}$ is a measurable set for all $a\in\mathbb{R}$ since $\{]-\infty,a]:a\in\mathbb{R}\}$ generates the Borel algebra of $\mathbb{R}$.
Jun
15
comment About the Wasserstein “metric”
By the way, here's some terminology explained: a $\it{transference\,\,plan}$ from $\mu$ to $\nu$ is a member of $\Pi(\mu,\nu)$. Intuitively, $d\pi(x,y)$ measures the amount of mass that $\pi$ transfers from $x$ to $y$ and $d(x,y)^{p}$ is the cost function. An optimal plan is such $\pi$ for which the infimum is reached in the definition of $W_{p}$, and for every other transference plan we have an inequality $\leq$. As mentioned above, given that $X$ is Polish guarantees the existence of optimal transference plans between any pair of Borel prob. measures.
Jun
15
revised About the Wasserstein “metric”
added one set for clarification.
Jun
15
revised About the Wasserstein “metric”
small typo
Jun
15
revised About the Wasserstein “metric”
added 11 characters in body
Jun
15
comment About the Wasserstein “metric”
A useful material as mentioned in my answer is the book of Cedric Villani "Topics in optimal transportation". Also one good source is Luigi Ambrosio's and Nicola Gigli's "User's guide to optimal transportation". It deals alot of nice results and detailed treatment of this topic. For instance, it considers the cases when $X$ is one of the following: Polish, geodesic metric space, Riemannian manifold, geodesic space with Alexandrov curvature (positive or nonpositive), and it defines a Metric Ricci curvature for a compact geodesic space $X$ through the study of its associated Wasserstein space.
Jun
15
answered About the Wasserstein “metric”