# Is the set of essentially bounded and not-necessarily-measurable functions a Banach space?

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 $$\|f\| = \esssup |f|$$?

• 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". – Cameron Williams Jan 30 at 13:17
• 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. – Shashi Jan 30 at 13:48
• @CameronWilliams Yes, thank you for that. – Harry Crimmins Jan 30 at 20:47
• @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. – Harry Crimmins Jan 30 at 20:51
• @HarryCrimmins yes indeed I missed it. – Shashi Jan 31 at 12:03

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.