Talk:Radon-Nikodym Theorem

Hadn't realised we already had trace $\sigma$-algebras set up here, (rather, I didn't know the proper name for a $\sigma$-algebra generated by intersections) and the part using the intersection measure might be a bit more aesthetically pleasing doing that instead. (avoiding having to play with the sum at the end) Will try to come back to it later, if anyone's passing by and wants to add it (as part of a Proof 2 with some refactoring I guess) feel free. Caliburn (talk) 15:38, 17 December 2021 (UTC)

Essential Uniqueness
I crated Characterization of Almost Everywhere Zero--Usagiop (talk) 20:27, 10 June 2022 (UTC)
 * This replicates Measurable Function Zero A.E. iff Absolute Value has Zero Integral it looks like. The proof of essential uniqueness is elaborate because the integrals may be infinite, so considering $\int \paren {g_1 - g_2} \rd \mu$ may not make sense. If you work with finite measures, it's as quick as you say. Caliburn (talk) 20:51, 10 June 2022 (UTC)
 * Why is my page replicates Measurable Function Zero A.E. iff Absolute Value has Zero Integral? I do not see any similarity.
 * $\int \paren {g_1 - g_2} \rd \mu$ may not make sense. No, unless $g_1 - g_2$ is integrable. But why are you taking about this? Did I say something to this type of integral?--Usagiop (talk) 20:57, 10 June 2022 (UTC)
 * Oh wait you are right, but I think it replicates some other theorem on here. (at least I remember doing something very similar) It may not have been proved. I will look at this in the morning. Caliburn (talk) 21:13, 10 June 2022 (UTC)
 * OK, it is possible. Anyway this type of claim should be stated somewhere else, not a part of Radon-Nikodym.--Usagiop (talk) 21:31, 10 June 2022 (UTC)
 * Are you saying this result should prove essential uniqueness? It doesn't seem to be sufficient. I think you will eventually want to write $\int \paren {g_1 - g_2} \rd \mu = \int g_1 \rd \mu - \int g_2 \rd \mu$ which doesn't work if $g_1$ and $g_2$ aren't integrable, even if $g_1 - g_2$ is. Caliburn (talk) 11:09, 11 June 2022 (UTC)
 * The claim in this proof is a specific case of Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal, just checked. Caliburn (talk) 11:23, 11 June 2022 (UTC)
 * You means for $A$ with $\map\mu A=\infty$, yes, you are right. There you probably need to make use of $\sigma$-finiteness.
 * Yes, the reference Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal is perfect.--Usagiop (talk) 13:07, 11 June 2022 (UTC)
 * You are correct about the mistake by the way - I think you want to find a cover that is finite under both $\mu$ and $\nu$. Taking the intersection of every pair of sets from both covers works for that. (it might be worth extracting this fact out and proving it separately, even) You're welcome to fill in this detail or I can get to it "in due course". Caliburn (talk) 14:02, 11 June 2022 (UTC)
 * You have already Lemma 2 for that.--Usagiop (talk) 18:43, 12 June 2022 (UTC)

I think instead of considering the intersection measure, we should instead restrict the measures, this seems to make everything work fine. I will get to this. Caliburn (talk) 15:18, 14 June 2022 (UTC)
 * I believe that fixes it, could you give it a once over? Caliburn (talk) 16:15, 14 June 2022 (UTC)

I am not sure if my ping worked on Definition talk:Restricted Measure, but pls see that page on how to resolve the omission. Thanks for your rigorous interrogation of this proof, by the way. We need much more of that here. Caliburn (talk) 18:19, 16 June 2022 (UTC)


 * Yes, I got an alert. I really do not care how you resolve it but I guess it is easier to define all $\mu _n, \nu _n$ as measures on $\struct {X,\Sigma}$. Here I saw such an approach Beppo Levi's Theorem ($\bigvalueat \mu {E_i}$ in proof).--Usagiop (talk) 18:36, 16 June 2022 (UTC)


 * I think it is most easily dealt with by using a function restriction, (my previous approach was imo far less elegant) so I think that is what I will do here. Caliburn (talk) 18:40, 16 June 2022 (UTC)