9.3. THE Lp SPACES 229
Proof: The second claim follows right away from the definition of ∥ f∥∞. If it is not
so, then µ ({ω : | f (ω)|> λ}) = 0 and so λ would be one of those constants M in thedescription of ∥ f∥
∞and ∥ f∥
∞would not really be the infimum of these numbers. Consider
the other claim. By definition,
µ
({ω : | f (ω)|> 1
n+∥ f∥
∞
})= 0
This is because there must be some essential upper bound M between ∥ f∥∞
and ∥ f∥∞+ 1
nsince otherwise, ∥ f∥
∞would not be the infimum. But
µ ({ω : | f (ω)|> ∥ f∥∞}) = ∪∞
n=1µ
({ω : | f (ω)|> 1
n+∥ f∥
∞
})and each of the sets in the union has measure zero. ■
Note that this implies that if ∥ f∥∞= 0, then f = 0 a.e. so f is regarded as 0. Thus,
to say that ∥ f −g∥∞= 0 is to say that the two functions inside the norm are equal except
for a set of measure zero, and the convention is that when this happens, we regard them asthe same function. If α = 0 then ∥α f∥
∞= 0 = 0∥ f∥
∞. If α ̸= 0, then M ≥ | f (ω)| implies
|α|M ≥ |α f (ω)| . In particular, |α|(∥ f∥
∞+ 1
n
)≥ |α f (ω)| for all ω not in the union of the
sets of measure zero corresponding to each ∥ f∥∞+ 1
n . Thus there is a set of measure zeroN such that for ω /∈ N,
|α|(∥ f∥
∞+
1n
)≥ |α f (ω)| for all n
Therefore, for ω /∈N, |α|∥ f∥∞≥∥α f∥
∞. This implies that ∥ f∥
∞=∥∥ 1
αα f∥∥
∞≤ 1|α| ∥α f∥
∞
and so ∥α f∥∞≥ |α|∥ f∥
∞also. This shows that this acts like a norm relative to multipli-
cation by scalars. What of the triangle inequality? Let Mn ↓ ∥ f∥∞
and Nn ↓ ∥g∥∞. Thus for
each n, there is an exceptional set of measure zero such that off this set Mn ≥ | f (ω)| and asimilar condition holding for g and Nn. Let N be the union of all the exceptional sets for fand g for each n. Then for ω /∈ N, the following holds for all ω /∈ N
Mn +Nn ≥ | f (ω)|+ |g(ω)| ≥ | f (ω)+g(ω)|
So take a limit of both sides and find that ∥ f∥∞+∥g∥
∞≥ | f (ω)+g(ω)| for all ω off a set
of measure zero. Therefore,
∥ f∥∞+∥g∥
∞≥ ∥ f +g∥
∞
Theorem 9.3.10 L∞ (Ω) is complete.
Proof: Let { fn} be a Cauchy sequence. Let N be the union of all sets where it isnot the case that | fn (ω)− fm (ω)| ≤ ∥ fn− fm∥∞
. By Proposition 9.3.9, there is such anexceptional set Mmn for each choice of m,n. Thus N is the countable union of these setsof measure zero. Therefore, for ω /∈ N,{ fn (ω)}∞
n=1 is a Cauchy sequence and so we letg(ω) = 0 if ω /∈ N and g(ω)≡ limn→∞ fn (ω) . Thus g is measurable. Also, for ω /∈ N,
|g(ω)− fn (ω)|= limm→∞| fm (ω)− fn (ω)| ≤ lim sup
m→∞
∥ fm− fn∥∞< ε
provided n is sufficiently large. This shows |g(ω)| ≤ ∥ fn∥∞+ ε, ω /∈ N so ∥g∥
∞< ∞.
Also it shows that there is a set of measure zero N such that for all ω /∈ N, for any ε > 0,|g(ω)− fn (ω)| < ε which means ∥g− fn∥∞
≤ ε . Since ε is arbitrary, this shows thatlimn→∞ ∥g− fn∥∞
= 0.■