proof of existence of the essential
supremum∗
gel†
2013-03-22 2:16:22
Suppose that (Ω, F, µ) is a σ-finite measure space and S is a collection of
measurable functions f : Ω → R̄. We show that the essential supremum of S
exists and furthermore, if it is nonempty then there is a sequence fn ∈ S such
that
ess sup S = sup fn .
n
As any σ-finite measure is equivalent to a probability measure, we may
suppose without loss of generality that µ is a probability measure. Also, without
loss of generality, suppose that S is nonempty, and let S 0 consist of the collection
of maximums of finite sequences of functions in S. Then choose any continuous
and strictly increasing θ : R̄ → R. For example, we can take

 x/(1 + |x|), if |x| < ∞,
1,
if x = ∞,
θ(x) =

−1,
if x = −∞.
As θ(f ) is a bounded and measurable function for all f ∈ S 0 , we can set
Z
0
α = sup
θ(f ) dµ : f ∈ S .
R
Then choose a sequence gn in S 0 such that θ(gn ) dµ → α. By replacing gn by
the maximum of g1 , . . . , gn if necessary, we may assume that gn+1 ≥ gn for each
n. Set
f = sup gn .
n
Also, every gn is the maximum of a finite sequence of functions gn,1 , . . . , gn,mn
in S. Therefore, there exists a sequence fn ∈ S such that
{f1 , f2 , . . .} = {gn,m : n ≥ 1, 1 ≤ m ≤ mn }.
∗ hProofOfExistenceOfTheEssentialSupremumi
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1
Then,
f = sup fn .
n
It only remains to be shown that f is indeed the essential supremum of S. First,
by continuity of θ and the dominated convergence theorem,
Z
Z
θ(f ) dµ = lim
θ(gn ) dµ = α.
n→∞
Similarly, for any g ∈ S,
Z
Z
θ(f ∨ g) = lim
θ(gn ∨ g) dµ ≤ α.
n→∞
It follows that θ(f ∨g)−θ(f ) is a nonnegative function with nonpositive integral,
and so is equal to zero µ-almost everywhere. As θ is strictly increasing, f ∨g = f
and therefore f ≥ g µ-almost everywhere.
Finally, suppose that g : Ω → R̄ satisfies g ≥ h (µ-a.e.) for all h ∈ S. Then,
g ≥ fn and,
g ≥ sup fn = f
n
µ-a.e., as required.
2