I saw a claim in some paper without proof. Because the formula is complicated, I wrote it in the attached file. I can not prove one part of the assertion. Please tell me how to prove it if you can understand.
Thank you
Please let me know if the following references/sites are helpful to you:
Solving Mathematical Problems by Terrance Tao
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Dennis
Dennis Mazur
Dear Dennis
Thank you for your reply.
All of your materials are very interesting.
It has a wide coverage and is too extensive for proof of my specific problem.
I will read them when I have time in the future.
After posting the question, I advanced a little.
Instead of proving directly that the limit value of any convergence sequence in the subsets E is contained in E, I have switched to prove that closure of some sub set of E coincides with E.
In this proof, I use the Lemma cited by the author, so I think I am getting closer to the correct answer.
Best regards
Mashiro Fujimoto
The Lemma proposed by the author guarantees that A is a subset of E. It is more a suggestion to use, to prove iii).
Dear Jorge
Thank you for your reply.
Prove (iii) does not need the Lemma.
For any $f \in L_0^ + ,\;s.t.\;f \le h,\;h \in E$, $f$ satisfies the condition of the definition of $E$.
$A \subset E$ is obvious because if $h \in E$, $h$ satisfies the condition of the definition of $E$ as ${h_n} = h$.
Best regards.
Mashiro Fujimoto
It is clear, that your question concerns a theorem in a paper by Walter Schachermayer, a professor in probability and quantitative finance in Vienna. My suggestion would be to ask professor Schachermayer for advise. By the way, in always pays to visit Schachermayer's homepage . Kind regards, J.W.N.
The Lemma is another formulation of that of Komlos, which can be used to prove the Doob-Meyer decomposition of a submartingale. This is a result that generalize the Bolzano-Wierstrass result for bounded sequences in R^n.
To prove that E is closed you just need to proof that the limit of every sequence is also in E, because of this Lemma you can always extract a subsequence that converges to something in A and so the limit of the sequence in E.
Dear Alejandro Santoyo
Thanks for your reply.
The attached file is my reply.
Best regard.
Masahiro Fujimoto
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