bio | website | |
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age | ||
visits | member for | 2 years, 1 month |
seen | 28 mins ago | |
stats | profile views | 82 |
Jan 22 |
revised |
Discrete measures and discrete kernels
added 49 characters in body |
Jan 22 |
comment |
Discrete measures and discrete kernels
I added the definition of support. |
Jan 22 |
revised |
Discrete measures and discrete kernels
Added definition of $\text{supp}\mu.$ |
Jan 22 |
comment |
Discrete measures and discrete kernels
Well, $K(x,\cdot)$ is also defined for $x\neq x_k.$ So the question is, if $K$ is still measurable in $x$ |
Jan 21 |
asked | Discrete measures and discrete kernels |
Nov 19 |
accepted | Relative interior and dense subsets |
Nov 19 |
comment |
Relative interior and dense subsets
@Kallus: you want to give this as an answer? |
Nov 18 |
comment |
Relative interior and dense subsets
But your second argument seems to be plausible... |
Nov 18 |
comment |
Relative interior and dense subsets
I think you are mixing up interior with relative interior. The relative interior of a non-empty convex set is always nonempty!!! |
Nov 18 |
revised |
Relative interior and dense subsets
edited body |
Nov 18 |
asked | Relative interior and dense subsets |
Jun 6 |
awarded | Enthusiast |
May 20 |
comment |
Existence of dominating measure for weak*-compact set of measures
What do you mean by universal property?Can you specify a bit more what you mean by compatibility in the first part? |
May 17 |
accepted | Existence of dominating measure for weak*-compact set of measures |
May 16 |
awarded | Commentator |
May 16 |
comment |
Existence of dominating measure for weak*-compact set of measures
This is really beautiful!The only thing I have still to think about is the existence of such a sequence $(\mathbb P_n){n\in\mathbb N}$... |
May 16 |
comment |
Existence of dominating measure for weak*-compact set of measures
@Davide : Does a weka*-compact substet necessarily have to be totally ordered? |
May 15 |
comment |
Existence of dominating measure for weak*-compact set of measures
@George Lowther : do you have a proof for this? So the answer given by Davide Giraudo is not true when the maps $Z$ are bounded? |
May 14 |
comment |
Existence of dominating measure for weak*-compact set of measures
@Gerald Edgar: "...the usual way to define it is to use continuous $Z$." Maybe you realized that we are working on measurable spaces and there is no notion of continuity... |
May 14 |
comment |
Existence of dominating measure for weak*-compact set of measures
I forgot boundedness! sorry |