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Apr
3
answered Axioms Conditional Probability
Apr
1
answered Decomposition of measures as a product measures
Apr
1
revised Convergence of nested converging measures
added 1 character in body
Apr
1
answered Convergence of nested converging measures
Apr
1
comment Equivalent Conditions for Weak Convergence of Signed Radon Measures
What does $|\nu_{k}| = 1$ mean? Is $|\cdot|$ some norm?
Mar
31
awarded  Guru
Mar
27
comment Application of Riesz Spaces (k/a $K$-Lineals or Vector Lattices ) to Mathematical-Economics?
See here for the first paper to use Riesz spaces in economics.
Mar
24
awarded  Good Answer
Mar
24
comment normed space is complete if and only if the closed unit ball is complete
A closed subset of a complete metric space is complete. For the other direction, take a Cauchy sequence and shift the values by some vector so that they all lie eventually in the closed unit ball.
Mar
22
comment Compactness, continuity and the discrete topology
1. Just let $f$ be continuous and bounded. 2. Even with non-Hausdorff topologies on a finite set, the resulting $\sigma$-algebra has a very simple form. Let $x\equiv y$ if $x$ and $y$ are members of exactly the same open sets. If $M$ is finit or even countably infinite, the Borel $\sigma$-algebra will consists of all arbitrary unions of $\equiv$-equivalence classes. This equivalence classes are like points. If you quotient out, you can pretend to be working in a discrete setting.
Mar
22
comment Compactness, continuity and the discrete topology
@Kolmin Are you sure you want $f$ to be linear?
Mar
22
answered $\mu$ is a $f$-invariant measure
Mar
22
comment Compactness, continuity and the discrete topology
How do you define weak convergence?
Mar
22
answered Definition of “deterministic coupling” [Villani]
Mar
14
comment Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?
@MikeBrown Uhm, yes it should.
Mar
14
comment Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?
@MikeBrown Formally, you show that the family $\{B\in\sigma(\mathcal{E}):B\in\sigma(\mathcal{C})\text{ for some countable subfamily }\mathcal{C}\subseteq\mathcal{E}\}$ is a $\sigma$-algebra. The elements of $\mathcal{E}$ are generators of $\sigma(E)$.
Mar
13
comment Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?
@MikeBrown The sets that are generated by some countable subfamily of generators.
Feb
24
awarded  Revival
Feb
15
awarded  Nice Question
Feb
8
awarded  Notable Question