The Lebesgue Integral

essentially bounded

A function is essentially bounded if it stays within some finite band — except possibly on a throwaway set of measure zero, which Lebesgue theory ignores. So a function that is 1000000 at a single point but at most 3 everywhere else still counts as essentially bounded by 3; the lone spike sits on a measure-zero set and is invisible to the measure.

Precisely, f is essentially bounded if there is a finite constant M such that |f(x)| less than or equal to M for almost every x. The smallest such M is the essential supremum of |f|, the genuine ceiling once measure-zero misbehavior is forgiven. This can be strictly smaller than the ordinary supremum, which would be ruined by a single tall point.

Essentially bounded functions are exactly the elements of L-infinity, with norm equal to the essential supremum. This is the right notion of boundedness inside the Lebesgue world: it is invariant under changing a function on a null set, fitting the rule that everything in L^p is defined only up to almost-everywhere equality.

Define f on [0,1] by f(x) = 5 for x irrational and f(x) = 100 for x rational. Since the rationals have measure zero, f is essentially bounded with essential sup 5, even though its ordinary sup is 100.

Essential sup 5, ordinary sup 100 — the spike on a null set is ignored.