Timeline for Problem on convergence in space of probability measures
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2015 at 13:46 | comment | added | user56932 | Thanks, understood that. So, if one has a sequence of convergent probability measure on $\overline{\phi(S)}$, then the corresponding probability measures (with zero measure on the boundary) converge in $\phi(S)$. $\phi$ is the homeomorphism. With this I am not able to verify the claim made in line 8 of the same page, same column. I think we need a uniform bound for $h$. | |
| Jan 13, 2015 at 12:56 | vote | accept | CommunityBot | ||
| Jan 13, 2015 at 12:09 | comment | added | Davide Giraudo | If $g$ is uniformly continuous and $x\in \overline S\setminus S$, take $x_n\in S$ such that $d(x,x_n)\to 0$. Since $g$ is uniformly continuous, the sequence $(g(x_n))$ is Cauchy and converges to a value $l$. It remains to show that this $l$ does not depend on the choice of the approximating sequence $(x_n)$. | |
| Jan 13, 2015 at 11:49 | comment | added | user56932 | Why "Such a function can be extended to a continuous (and bounded) function on $\overline S$, denoted $\overline g$." ? That is my main doubt. | |
| Jan 13, 2015 at 10:27 | history | answered | Davide Giraudo | CC BY-SA 3.0 |