Let $u_n, u \in L^{\infty}(E)$ and let $u_n \rightarrow u$ almost everywhere. Is it true that $$ \DeclareMathOperator{\esssup}{ess sup} \sup_{m \geq n} \, \esssup_E |u_n - u_m| = \esssup_E \, \sup_{m \geq n} |u_n - u_m| \qquad ?$$
2026-04-05 14:47:44.1775400464
$\sup_{m \geq n} \, \operatorname{ess sup} |u_n - u_m| = \operatorname{ess sup} \, \sup_{m \geq n} |u_n - u_m|$
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No. Take $u_n(t)=\begin{cases}0&\text{, if }t\neq n\\1&\text{, if }t=n\end{cases}$.