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Feb 26, 2019 at 17:51 comment added coudy @Sinusx $f$ is not assumed to be integrable.
Feb 22, 2019 at 23:43 comment added Viktor B Isn't it a consequence of the dominated convergence theorem?
Jul 26, 2018 at 18:01 comment added coudy @YemonChoi Indeed, thanks for pointing the correct reference.
Jul 23, 2018 at 4:26 comment added Yemon Choi Are you sure Joel's article was in the AMM? According to arxiv.org/abs/1412.7702 it was published in the Irish Mathematical Society's Bulletin
Jun 5, 2018 at 19:08 history edited Piotr Hajlasz CC BY-SA 4.0
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May 1, 2018 at 16:20 comment added BCLC @lcv DCT has integrable $g$ rather than nonnegative $f$ right?
Apr 25, 2018 at 16:11 comment added Ovi I am an undergrad just familiar with the Lebesgue measure, not with measure spaces. I am surprised that about your example; I haven't had time to digest it, but I was under the impression that the D.C.T. and Fatou;s lemma imply the result in your post. If $f$ is integrable, then it is a dominating function and D.C.T applies. If not, then $\int f = \infty$ and by Fatou's lemma $\int f_n \to \infty$. I am guessing that this is true somehow only with the Lebesgue measure, and not in other measure spaces? Othewise I don't understand how your example would be possible.
Apr 1, 2018 at 20:15 comment added coudy @lcv this is an extension of the monotone convergence theorem. The function is assumed to be non-negative.
Apr 1, 2018 at 19:41 comment added lcv Is this an extension of the monotone convergence or the dominated convergence theorem?
Mar 30, 2018 at 13:48 comment added Arnaud D. According to this MSE question, this theorem appears in at least one textbook : math.stackexchange.com/questions/2713020/…
S Mar 29, 2018 at 19:43 history answered coudy CC BY-SA 3.0
S Mar 29, 2018 at 19:43 history made wiki Post Made Community Wiki by coudy