# Exchanging limit with supremum

I have showed that for all ϵ>0\epsilon > 0 and for all ff in {f}\{f\},
there exists N′N’ such that for all n≥N′n \ge N’ ∫|f|I{f≥N}<ϵ/2\int |f| I_{\{f \ge N \}} < \epsilon/2. I would like to claim that limN→∞supf∫|f|I{f≥N}=0\lim_{N \rightarrow \infty} \sup_f \int |f| I_{ \{f\ge N\}} =0. But in so doing, I am not so sure if I am allowed to claim such. I want to make it crystal clear but something is iffy. Can you give me a way to see it clearly? =================      You cannot deduce your claim just like that. You need others assumptions, because from "pointwise" (in ff) convergence to 00 doesn't give you "uniform" convergence to 00. – anonymus Oct 21 at 0:47 ================= =================