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 Feb1 comment If $x_n \rightarrow 0$ and $\{y_n\}$ is a bounded sequence, then $x_ny_n \rightarrow 0$. Just be careful of your $\alpha$. It must be a point so that $\alpha \in \mathbb{R}^+$. And yes, the same $N$ works because $y_n$ is bounded independent of whether $x_n$ converges or not. Jan12 accepted Part of Fubini's Theorem with almost everywhere Jan12 comment Part of Fubini's Theorem with almost everywhere This question was already asked math.stackexchange.com/questions/1092685/… Jan12 accepted Tonelli and Fubini and almost everywhere Jan12 asked Part of Fubini's Theorem with almost everywhere Jan6 asked Riemann and Lebesgue improper integral Proof Jan6 revised Tonelli and Fubini and almost everywhere added 65 characters in body Jan6 comment Tonelli and Fubini and almost everywhere I had forgotten about this theorem. Thank you. Jan6 asked Tonelli and Fubini and almost everywhere Jan5 accepted Folland 2.36 portion of proof Jan5 comment Folland 2.36 portion of proof This is the conclusion I arrived at as well. Thank you! Jan5 asked Folland 2.36 portion of proof Dec22 accepted Question about simple functions as described in Folland's Real Analysis Dec22 comment Question about simple functions as described in Folland's Real Analysis I'm a fool. Thank you so much. Dec22 comment Question about simple functions as described in Folland's Real Analysis Yes, I understand that part of the proof. My question is why does this proof work for the more general case. I'm not seeing why $\phi$ being simple and 0 almost everywhere imply that it's integral is 0 since this is something we are trying to prove in the first place. Dec21 asked Question about simple functions as described in Folland's Real Analysis Dec17 awarded Yearling Dec17 asked Improper Riemann and Lebesgue Integrals Nov19 comment representing intervals as infinite intersectiom or union Yes, but I think my issue is I think of these as limits which I'm sure I wrong. Nov19 asked representing intervals as infinite intersectiom or union