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Egorov's Theorem: Proof Explained

Egorov's Theorem states that if a sequence of functions converges pointwise almost everywhere to another function on a finite measure space, then the convergence can be made uniform after removing a set of arbitrarily small measure. The proof shows that the space can be written as a nested intersection of sets where the functions converge uniformly. It then uses properties of measure and continuity to show there exists a set whose removal makes the convergence uniform everywhere else, and whose measure can be made arbitrarily small.

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454 views2 pages

Egorov's Theorem: Proof Explained

Egorov's Theorem states that if a sequence of functions converges pointwise almost everywhere to another function on a finite measure space, then the convergence can be made uniform after removing a set of arbitrarily small measure. The proof shows that the space can be written as a nested intersection of sets where the functions converge uniformly. It then uses properties of measure and continuity to show there exists a set whose removal makes the convergence uniform everywhere else, and whose measure can be made arbitrarily small.

Uploaded by

peter
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Egorov’s Theorem, a detailed proof.

Theorem: Let (X, M, µ) be a measure space with µ(X) < ∞. Let {fn } be a sequence of measurable
functions on X and let f be a measurable function on X. Assume that fn → f a.e. pointwise. Then for

any ϵ > 0, there exists a measurable set D of X, such that µ(D) < ϵ and fn → f uniformly on X − D.
Proof.

As fn → f pointwise a.e. on X, without loss of generality, we can just assume that fn → f pointwise
on X.
From the ϵ-N definition of fn (x) → f (x), and note that fn → f on X, for any ϵ > 0, we have

∪ ∩
X= {x : |fn (x) − f (x)| < ϵ}.
N ∈N≥1 n>N

Use En,ϵ to denote {x : |fn (x) − f (x)| < ϵ}. Note that 1/k → 0 as k → ∞, we then translate the ϵ-N
language of pointwise convergence on X to

∪ ∩
X= En,1/k for all k ∈ N≥1 .
N ∈N≥1 n>N

Note that
∩ ∩
En,1/k ⊂ En,1/k ⊂ · · · .
n>1 n>2

By the continuity of measures, we have


( )
∩
µ(X) = lim µ En,1/k .
N →∞
n>N

As µ(X) < ∞, it follows that (as no such case as ∞ − ∞ is involved)

∩ ∩ ∩
µ(( En,1/k )c ) = µ(X − En,1/k ) = µ(X) − µ( En,1/k ) → 0 as N → ∞.
n>N n>N n>N

1
Then for all k ∈ N≥1 , we can choose Nk ∈ N≥1 , such that

∩
µ(X − En,1/k ) < ϵ/2k .
n>Nk

∩
Use Xk to denote n>Nk En,1/k . As µ(X − Xk ) < ϵ/2k for all k, and note that

( )c
∩ ∩ ∪ ∪
X− Xk = Xk = Xkc = (X − Xk ),
k≥1 k≥1 k≥1 k≥1

it follows that ( )
∩ ∑ ∑
µ X− Xk ≤ µ(X − Xk ) ≤ ϵ/2k = ϵ.
k≥1 k≥1 k≥1
∩
Let D = X − k≥1Xk . Then D is measurable and µ(D) < ϵ. It remains to show that fn → f
∩
uniformly on X − D = k≥1 Xk .
∩ ∩
In fact, for any integer m ≥ 1, note that k≥1 Xk ⊂ Xm . Thus for any x ∈ k≥1 Xk , we also have
x ∈ Xm . According to the definition of Xm , for any n > Nm , we have

|fn (x) − f (x)| < 1/m,

which finishes the proof. Q.E.D.

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