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Suppose that $(\Omega, \mathcal{F}, \mu)$ is a probability space and for $f : \Omega \to \mathbb{R}$ define $$\DeclareMathOperator{esssup}{ess\,sup} \esssup f = \inf_{A : \mu(A) = 0} \sup_{x \notin A} f(x). $$ To me it seems that a short elementary argument shows that for measurable $f$ this agrees with the usual definition of the essential supremum, yet there is nothing in the definition that requires $f$ to be measurable.

Is it true that the set of equivalences classes of a.e. equal, bounded $\mathbb{R}$-valued functions on $\Omega$ is a Banach space when equipped with the norm $\|f\| = \esssup |f|$?

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    $\begingroup$ To answer the question in the title: no since you could add a function and its negative to get the zero function which is measurable. The title probably should read "not necessarily measurable". $\endgroup$ – Cameron Williams Jan 30 at 13:17
  • $\begingroup$ What I see happening is that you probably can find a function for which you cannot determine its norm... Intuitively, you can only distinguish measurable functions from each other. I'm not sure tho. $\endgroup$ – Shashi Jan 30 at 13:48
  • $\begingroup$ @CameronWilliams Yes, thank you for that. $\endgroup$ – Harry Crimmins Jan 30 at 20:47
  • $\begingroup$ @Shashi Well the given definition of ||f|| is well-defined for bounded f, so I don't see a problem there. What I am suggesting is that you can then quotient by the kernel of || || and get a Banach space. $\endgroup$ – Harry Crimmins Jan 30 at 20:51
  • $\begingroup$ @HarryCrimmins yes indeed I missed it. $\endgroup$ – Shashi Jan 31 at 12:03
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The set $B$ of ALL bounded functions is a Banach space for the sup norm. The subset $N$ formed by the functions vanishing off a null set is a closed subspace. Your space is exactly the quotient space $B/N$ with the quotient norm, hence, yes, it is a Banach space.

The reason it isn't studied in the literature is that there is not much one can say about it. In particular there is no well defined linear functional given by integration.

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  • $\begingroup$ This is the answer I was looking for. Thanks $\endgroup$ – Harry Crimmins Jan 30 at 21:08

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