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Apr 30, 2022 at 14:32 history edited kjetil b halvorsen
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Jun 20, 2019 at 18:50 comment added thegain Thanks for the answer. Actually, in my case all the $\nu_n$ are absolutely continuous with respect to $\mu$, and I can even show that all the weak cluster points of the sequence $\nu_n$ also are.
May 31, 2019 at 12:34 comment added R W Keep in mind that although the sequence $\nu_ n$ weakly converges to $\nu$, nothing precludes the measures $\nu_n$ from being mutually singular with $\mu$.
May 31, 2019 at 8:15 review First posts
May 31, 2019 at 8:56
May 31, 2019 at 8:11 history asked thegain CC BY-SA 4.0