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Jun 8, 2020 at 19:00 comment added LSpice The name of the linked paper by Kelley: Decomposition and representation theorems in measure theory.
Jun 8, 2020 at 18:17 comment added Dmitri Pavlov @ogogmad: This is Theorem 2 in Kelley's paper that is linked in the answer.
Jun 8, 2020 at 18:16 history edited Dmitri Pavlov CC BY-SA 4.0
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Jun 8, 2020 at 18:10 comment added wlad Can you link to a reference to your Hahn-Jordan decomposition theorem?
Jun 8, 2020 at 17:58 comment added Dmitri Pavlov @ogogmad: First of all, the Lebesgue decomposition theorem does not require σ-finiteness or any other conditions (see the nLab). Secondly, as I already explained, it has nothing to do with the Hahn–Jordan decomposition, which decomposes a signed measure into its positive and negative components.
Jun 8, 2020 at 17:52 comment added wlad Wikipedia says that Lebesgue decomposition requires that the measures $\mu$ and $\nu$ be $\sigma$-finite. This can surely be relaxed, but it doesn't hold unconditionally
Jun 8, 2020 at 17:49 comment added Dmitri Pavlov @ogogmad: No. I am not even sure what you are trying to say here, since the Lebesgue decomposition theorem is trivial in the case of absolute continuity, which is a precondition for the Hahn–Jordan theorem.
Jun 8, 2020 at 17:32 comment added wlad Sounds like the Lebesgue decomposition theorem, no?
Jun 8, 2020 at 15:32 comment added Dmitri Pavlov @ogogmad: It's a different Hahn–Jordan decomposition theorem (one measure instead of two). I added a clarification.
Jun 8, 2020 at 15:30 history edited Dmitri Pavlov CC BY-SA 4.0
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Jun 8, 2020 at 6:47 comment added wlad The Hahn-Jordan decomposition theorem is stated on Wikipedia without any preconditions
Jun 8, 2020 at 1:20 review Suggested edits
Jun 8, 2020 at 2:03
Jun 7, 2020 at 20:27 history edited Dmitri Pavlov CC BY-SA 4.0
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Jun 7, 2020 at 20:07 history answered Dmitri Pavlov CC BY-SA 4.0