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Sep 26, 2019 at 3:06 vote accept jpv
Sep 25, 2019 at 15:45 answer added Nate Eldredge timeline score: 4
Sep 25, 2019 at 14:58 comment added jpv Thanks. I was not aware of universally measurable sets. I will look for some references.
Sep 25, 2019 at 14:56 comment added Nate Eldredge If $(\Omega, \Sigma)$ is standard Borel then $X(A)$ is in fact universally measurable (though not necessarily Borel), and therefore it does contain a Borel set of measure 1.
Sep 25, 2019 at 14:55 comment added jpv Any assumptions on $(\Omega, \Sigma,P)$ are fine at this point for me.
Sep 25, 2019 at 14:53 comment added Nate Eldredge Do we have any assumptions on $(\Omega, \Sigma)$, e.g. standard Borel?
Sep 25, 2019 at 14:48 history asked jpv CC BY-SA 4.0