Skip to main content
8 events
when toggle format what by license comment
Jan 16, 2021 at 15:59 history edited YCor CC BY-SA 4.0
formatting
May 25, 2014 at 11:08 comment added Michael Greinecker You can find a proof in Stochastic Optimal Control: The Discrete-Time Case by Bertsekas and Shreve, where this is Proposition 7.25.
May 22, 2014 at 18:24 vote accept TV2323
May 22, 2014 at 1:11 answer added Nate Eldredge timeline score: 6
May 22, 2014 at 0:51 answer added Alexander Shamov timeline score: 6
May 22, 2014 at 0:49 comment added Nate Eldredge To show that $\mu \mapsto \mu(A)$ is measurable for every Borel $A$, try a $\pi$-$\lambda$ or monotone class argument. Let $\mathcal{L}$ be the set of all $A \subset X$ such that $\mu \mapsto \mu(A)$ is measurable, and let $\mathcal{P}$ be the closed subsets of $X$. Then it is easy to check that $\mathcal{P}$ is closed under finite intersections, $\mathcal{L}$ is a $\lambda$-system, and $\mathcal{P} \subset \mathcal{L}$. The conclusion is that $\mathcal{L} \supset \sigma(\mathcal{P})$, i.e. the Borel sets.
May 21, 2014 at 21:52 review First posts
May 21, 2014 at 23:14
May 21, 2014 at 21:34 history asked TV2323 CC BY-SA 3.0