Page 320 - DMTH401_REAL ANALYSIS

Real Analysis

Notes

Notes Above Lemma 4 is false if the condition m(F ) < ¥ is missing.
1
Now, we are ready to prove Littlewood’s 2nd Principle.
Proof: Let  > 0 be given. Our proof is divided into two steps.
Step I: Assume m  f  M for some m, M  .
We divide [m, M] into n subintervals such that the length of each subinterval is less than .
Symbolically, we take the partition points as follows:
m = y < y <  < y = M with y – y <for 1  i  n.
0 1 n i i–1
Let E = {x  E : m  f(x)  y }, E = {x  E : y < f(x)  y },..., E = {x  E : y < f(x)  M }. Now, take
1 1 2 1 2 n n–1
 = y  + y  +  + y  . Since E , E ,..., E are all measurable (why?),  is simple and satisfies
1 E 1 2 E 2 n E n 1 2 n
the inequality |f – | <  with no exceptions.
Step II: General case.
We let
F = {x  E : |f(x)|  n}.
n
Then F  F  . Note that m(F )  m(E) < ¥ and m(F ) = 0 by assumption, apply Lemma 4 there
1 2 1 ¥
exists N   such that
m(F ) < .
N
Now, let f* = (–N  f)  N, then f = f* on E except on a set of measure less than . From the result
of Step I, there is a simple function  such that |f* – | <  on E. Hence
|f – | <  on E except on a set of measure less than .

Corollary: There is a sequence of simple functions  such that   f pointwisely almost
n n
everywhere on E. If E = [a, b], there are also sequence of step functions and sequence of continuous
functions converging to f pointwisely almost everywhere on [a,b].
n
Proof: Applying Littlewood’s 2nd Principle to  = 1/2 , there are simple functions  and sets A
n n
n
with m(A ) < 1/2 such that
n
1
|f –  |< on E\A .
n n n
2
Let A = lim A :=  ¥ k 1 ( ¥ n k A ), then m(A) = 0 (why?). The proof is completed by noting that   f
n = = n n
pointwisely on E\A.

Notes In fact, the sequence  can be chosen so that   f pointwisely everywhere on E.
n n
2
For example, we can first divide the interval [–n, n] into 2n subintervals such that each
subinterval has length 1/n, i.e. choose
–n = y < y <  < y 2 = n
0 1 2n
such that y – y = 1/n for all i. Then let
i i–1
ì y i if y  f(x) < y i 1 for some i
+
i
ï
 (x) = í n if f(x)  n
n
ï - n if f(x) < - n
î
314 LOVELY PROFESSIONAL UNIVERSITY