3.

9 Egoroff’s Theorem 105

3.9 Egoroff’s Theorem

We know that pointwise convergence of functions does not imply uniform con-
vergence, and likewise pointwise a.e. convergence does not imply L∞ norm
convergence. A standard counterexample is the Shrinking Boxes of Exam-
ple 3.62: χ[0, k1 ] converges pointwise a.e. to the zero function, but the conver-
gence is not uniform. The functions χ[0, k1 ] are not continuous, but this is not
the issue. For example,


 0, x ≤ 0,

1

linear, 0 < x < 2k ,



fk (x) = 1
1, x = 2k ,

linear, 2k < x < k1 ,
1





x ≥ k1 .

0,


is a continuous function and fk → 0 pointwise, but fk does not converge

uniformly to the zero function (see the illustration in Figure 3.1).

1.0
f10 f2
0.8

0.6

0.4

0.2

0.0
0.0 0.2 0.4 0.6 0.8 1.0

Fig. 3.1. Graphs of the functions f2 (dashed) and f10 (solid).

However, if we allow ourselves to reduce the domain of these functions,

then we can find a subset on which we have uniform convergence. In particu-
lar, if 0 < δ < 1 then the functions fk converge uniformly to 0 on the interval
[δ, 1]. Egoroff’s Theorem essentially states that this example is typical, as long
as we are dealing with a finite measure space. So it is important in the ex-
ample above that the region of interest (the interval [0, 1]) has finite measure,
but according to Egoroff, whenever we have pointwise a.e. convergence of a
sequence of functions in a finite measure space, the sequence will actually
converge uniformly on a “large” part of the domain.
Theorem 3.68 (Egoroff ’s Theorem). Let (X, Σ, µ) be a finite measure
space. If fk , f : X → C are measurable functions such that fk → f point-
wise a.e., then for every ε > 0 there exists a measurable set E ⊆ X such
that
c 2011 by Christopher Heil
106 3 Measurable Functions

(a) µ(E) < ε, and

(b) fk converges uniformly to f on E C , i.e.,
 
lim sup |f (x) − fk (x)| = 0.
k→∞ x∈E
/

Proof. Let Z be the set of measure zero consisting of all points x ∈ X such
that fk (x) does not converge to f (x). For each k, n ∈ N, define the measurable
sets
∞ n
S 1o ∞
T
Ek (n) = |f − fm | ≥ and Zn = Ek (n).
m=k n k=1

Fix n, and suppose that x ∈ Zn . Then x ∈ Ek (n) for every k, so for each k
there must exist some integer m ≥ k such that |f (x) − fm (x)| > n1 . Therefore
fk (x) does not converge to f (x) as k increases, so this point x belongs to Z.
This shows that
Zn ⊆ Z,
and therefore µ(Zn ) = 0 by monotonicity. With n fixed, the sets Ek (n) are
nested decreasing, and their intersection is Zn by definition. Therefore, it
follows from continuity from above that


∀ n ∈ N, lim µ Ek (n) = µ(Zn ) = 0. (3.13)
k→∞

Fix ε > 0. Applying equation (3.13), for each integer n there is some
integer kn > 0 such that

ε
µ Ekn (k) < n .
2
Define ∞
S
E = Ekn (n).
n=1

Subadditivity implies that µ(E) ≤ ε. Further, if x ∈ / Ekn (k) for

/ E then x ∈
any n, so |f (x) − fm (x)| < n1 for all m ≥ kn .
In summary, µ(E) ≤ ε and for each n ∈ N there exists an integer kn > 0
such that
1
m ≥ kn =⇒ sup |f (x) − fm (x)| ≤ .
x∈E
/ n
This says that fk converges uniformly to f on E C . ⊓
⊔

Additional Problems

3.23. Let (X, Σ, µ) be a finite measure space, and let fk , f : X → R be mea-

m
surable function on X. Show that if fk → f pointwise a.e., then fk → f.
 3.9 Egoroff’s Theorem 107

3.24. Let D be a Lebesgue measurable subset of Rd such that |D| < ∞. Show
that if fk , f : D → C are measurable and fk → f a.e. on D, then there exists
a closed set F ⊆ D such that |D\F | < ε and fk → f uniformly on F.

3.25. Suppose that µ is a σ-finite measure on X, and fk , f : X → C are

measurable functions such that fk → f a.e. Show that there exist measurable
sets En ⊆ E such that


(a) µ X \ En = 0, and
S

(b) fk → f uniformly on each set En .