Page 344 - DMTH401_REAL ANALYSIS

Real Analysis

Notes lim sup f : X ® 
n
is the function defined by
(lim sup f ) (x) : = lim sup(f (x)) at each x  X.
n n
 If {f } is a sequence of measurable functions, then the functions
n
sup f , inf f , lim sup f , and lim inf f
n n n n
are all measurable. If the limit lim f (x) exists at each x  X, then the limit function lim f
n®¥ n n
is measurable. For instance, if the sequence {f } is monotone, that is, either non-decreasing
n
or non-increasing, then lim f is everywhere defined and it is measurable.
n
 A function is measurable if and only if it is the limit of simple functions. Moreover, if the
function is non-negative, the simple functions can be taken to be a non-decreasing sequence
of non-negative simple functions.

28.5 Keywords

Limits Preserve Measurability: If {f } is a sequence of measurable functions, then the functions
n
sup f , inf f , lim sup f , and lim inf f
n n n n
are all measurable.
Characterization of Measurability: A function is measurable if and only if it is the limit of
simple functions. Moreover, if the function is nonnegative, the simple functions can be taken to
be a non-decreasing sequence of nonnegative simple functions.

Uniformly Convergent: Every convergent sequence of real-valued measurable functions is almost
uniformly convergent.
Egorov’s Theorem: On a finite measure space, a.e. convergence implies a.u. convergence for real-
valued measurable functions.

28.6 Review Questions

1. Let A , A ,... be measurable sets and put
1 2
¥ ¥ ¥ ¥
lim sup A :=   A and lim inf A :=   A .
n k n k
=
n 1 k n n 1 k n
=
=
=
Let f and f be the characteristic functions of limsup A and liminf A , respectively, and for
n n
each n, let f be the characteristic function of A . Prove that
n n
f = lim sup f and f = lim inf f .
n n
(i) First prove the theorem for simple functions. Suggestion: Let f be a simple function
and write f = å N k=1 a c where X =  N k 1 A , the a ’s are real numbers, and the A ’s are
k
=
A
k
k k k
pairwise disjoint measurable sets. Given  > 0, there is a closed set C   with
n
k
m(A \C ) < /N (why?). Let C =  N k 1 C .
k k = k
(ii) We now prove Luzin’s theorem for non-negative f. For nonnegative f we know that
f = lim f where each f , k  , is a simple function. By (i), given  > 0 there is a closed
k k
k
set C such that m(X\C ) < /2 and f is continuous on C .
k k k k
Let K =  ¥ k 1 C . Show that m(X\K ) < . Use Egorov’s theorem to show that there
1 = k 1
exists a set K  K with m(K \K ) <  and f ® f uniformly on K . Conclude that f is
2 1 1 2 k 2
continuous on K .
2
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