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How can I prove that the essential supremum is itself an essential upper bound?

i.e. For an essentially bounded function, in the relation $S=\inf\{M: |f(x)| ≤ M \mbox{ a.e.}\}$, can the infimum be taken as a minimum?

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Yes, if $S$ is finite, then for a fixed $n$, take $N_n$ of measure $0$ such that on its complement, $|f(x)|\leq S+\frac 1n$. Then take $N:=\bigcup_{n\geq 1}N_n$. $N$ is of measure $0$, and on its complement $|f(x)|\leq S$.

But we may not have $|f(x)|, for example if we take $f$ constant.

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    Yeah @t.b. you are right, I already changed it in the original post.2012-04-01