Page 342 - DMTH401_REAL ANALYSIS

Real Analysis

Notes In particular, since products and reciprocals of measurable functions are measurable, whenever
the reciprocal is well-defined, it follows that quotients of measurable functions are measurable,
whenever the denominator is nonvanishing.

28.3 Littlewood’s Third Principle

We finally come to the third of Littlewood’s principles, which is
Every convergent sequence of [real-valued] measurable functions is nearly uniformly
convergent, or, more precisely, in the words of Lebesgue who in 1903 stated this principle as.
Every convergent series of measurable functions is uniformly convergent when certain sets of
measure  are neglected, where  can be as small as desired.

Lebesgue here is introducing the idea which is nowadays called “convergence almost uniformly.”
A sequence {f } of measurable functions is said to converge almost uniformly (or “a.u.” for short)
n
to a measurable function f, denoted by
f ® f a.u.,
n
c
if for each  > 0, there exists a measurable set A such that (A) <  and f ® f uniformly on A = X\A.
n
c
As a quick review, recall that f ® f uniformly on A means that given any  > 0,
n
c
|f (x) – f(x)| < , for all x  A and n sufficiently large.
n
c
Note that f (x) and f(x) are necessarily real-valued (cannot take on ±¥) on A . Therefore, Lebesgue
n
is saying that
Every convergent sequence of real-valued measurable functions is almost uniformly convergent.
The following theorem, although stated by Lebesgue in 1903, is named after Dimitri Fedorovich
Egorov (1869-1931) who proved it in 1911[34].
Theorem 4: Egorov’s Theorem
On a finite measure space, a.e. convergence implies a.u. convergence for real-valued measurable
functions. That is, any sequence of real-valued measurable functions that converges a.e. to a real-
valued measurable function converges a.u. to that function.

Proof: Let f, f , f , f ,... be real-valued measurable functions on a measure space X with (X) < ¥,
1 2 3
and assume that f = lim f a.e, which means there is a measurable set A  X with (X \A) = 0 and
n
f(x) = lim f (x) for all x  A. We need to show that f ® f a.u.
n®¥ n n
Step 1: Given ,  > 0 we shall prove that there is a measurable set B  X and an N   such that
(3.3) (B) <  and for x  B , c |f(x) – f (x)| <  for all n > N.
n
Indeed, for each m  , put

B :=  {x X; f(x) f (x) ³  }

-
m n
³
n m
Notice that each B is measurable and B  B  B  . Also, since for all x  A, we have f (x) ®
m 1 2 3 n
f(x) as n ® ¥, it follows that if x  A, then |f(x) – f (x)| <  for all n sufficiently large. Thus, there
n
is an m such that x  B , and so, x  A  x  B for some m. Taking contrapositives we see that
m m
x  B for all m  x  A, which is to say,
m
¥
 B  X \A.
m
=
m 1
Thus,  ( ¥ m 1 B )  (X\A) = 0 and therefore, since X is a finite measure space, by continuity of
m
=
measures (from above), we have
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