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I'm sure this must've been asked before here but I couldn't find it. I'm trying exercise 11.9 from Le Gall's "Measure theory, Probability, and Stochastic processes". The objective is to prove what I wrote in the title, when $X,Y\in L^1$. It starts by asking to prove the case when they're $L^2$, and then prove that for $a>0$ $$\mathbb E[X\mid X\wedge a]\wedge a=X\wedge a$$ I have managed to do both of these things. Then it follows by asking to prove that $$\mathbb E[X\wedge a\mid Y\wedge a]=Y\wedge a$$ which I haven't managed; I do see how to conclude from there that $X=Y$ but I'm stuck on that step. The book has teh hint to prove $\mathbb E[X\wedge a\mid Y\wedge a]\leq Y\wedge a$ which I deed by using that $\min(\cdot,a)$ is concave and using that $\mathbb E[\mathbb E[\cdot\mid Y\wedge a]\mid Y]=\mathbb E[\mathbb E[\cdot\mid Y]\mid Y\wedge a]$, but I don't see where to go from there; I've tried to find a clever conditioning, or to decompose $X=X1_{[X<a]}+X1_{[X\geq a]}$ but haven't gotten anything. Any help is appreciated! I feel like I'm missing some crucial understanding about conditional expectation, as my intuition is very poor...

EDIT: I was made aware in the comments that this has been asked before here, I understand all proofs in that link but this one, which uses the approach I'm trying to use. To me it seems like they only prove the same inequality I proved. I would like to see someone show me how to go from that inequality to equality; to see how the book intended me to solve it

Bruno Andrades
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    This has been answered earlier. It is, in fact, a popular question. – Kavi Rama Murthy Jun 18 '25 at 04:37
  • Also, you have to first reduce the proof to non-negative case. Otherwise, these ideas won't work. – Kavi Rama Murthy Jun 18 '25 at 04:39
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    See: https://math.stackexchange.com/questions/666843/if-exy-y-almost-surely-and-eyx-x-almost-surely-then-x-y-almost-surel – Kavi Rama Murthy Jun 18 '25 at 07:23
  • Thanks! @KaviRamaMurthy – Bruno Andrades Jun 18 '25 at 11:03
  • @KaviRamaMurthy in this answer, they use the same method as the book suggest, but to me it seems like they don't finish the proof, as they just seem to prove the same inequality I proved, am I missing something obvious? https://math.stackexchange.com/a/3605516/578498 – Bruno Andrades Jun 18 '25 at 11:12
  • The proof looks very clear .If $Z$ is a non-negative random variable and $EZ=0$ then $Z=0$ a.s. – Kavi Rama Murthy Jun 18 '25 at 11:17
  • @KaviRamaMurthy I'm sorry if I wasn't clear, I meant the last answer, who tries the idea the book suggested... Both the book and said answer seem to suggest that if $X,Y\in L^1$ you don't need to assume they're non-negative – Bruno Andrades Jun 18 '25 at 11:21
  • I am not saying that $X,Y \ge 0$ is necessary for the result to be true. I was only talking about the proof. It is easy to reduce to this special case , and that makes the proof simpler. – Kavi Rama Murthy Jun 18 '25 at 11:24
  • @KaviRamaMurthy I see, then it's not clear to me how that is what they're using, are we talking about the same answer? I'm talking about the answer linked above – Bruno Andrades Jun 18 '25 at 11:25

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