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May 7, 2022 at 12:35 history edited Kurome CC BY-SA 4.0
added 81 characters in body
Apr 21, 2022 at 7:27 comment added Jochen Wengenroth The most natural thing to try is to define $\tilde\mu(M)=\mu(B)$ if $B\in\mathscr R$ is such that $M\Delta B$ is $\mu$-null. The first thing to check is that this is well-defined which should follow from $B\Delta C=(B\Delta M)\Delta (C\Delta M)$ (this is best seen using the group homomorphism $(\mathscr P(X),\Delta)\to (\{0,1\}^X,\oplus)$, $M\mapsto 1_M$ (the indicator function) where $1\oplus 1=0$).
Apr 20, 2022 at 22:13 history edited YCor CC BY-SA 4.0
removed capitals from title
Apr 20, 2022 at 21:55 history edited LSpice CC BY-SA 4.0
Caratheodory -> Carathéodory
Apr 20, 2022 at 21:53 comment added LSpice Idiom: something is taken up a notch, not taken to a notch (although that's arguably perfectly logical phrasing). \\ The definition of $\mu$-null does not involve $A$.
Apr 20, 2022 at 15:42 comment added Kurome Hi, you're right. It's quite unproductive, and my frustration got the best of me. I'll edit after I finish with my tasks right now.
Apr 20, 2022 at 15:36 comment added John Wiltshire-Gordon Hi Kurome, I think this would be a better question if you replace the rant with a short motivation of delta rings (including any benefits they offer over sigma algebras), even if you don't find that motivation completely convincing.
Apr 20, 2022 at 15:20 comment added Kurome If anybody knows of a more elementary text that discusses exactly this, can someone point me to it?
Apr 20, 2022 at 15:09 history asked Kurome CC BY-SA 4.0