Page 376 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
+
A ò (f g) = ò A f + A ò g
A ò  =  A ò f
f
Proof:

(a) This is clearly true, for if j Î B (E) and j £ f on A, then j £ g on A so ò j £ A ò g by definition
0 A
of ò g . Taking supremum over all such j’s, we get ò f £ A ò g .
A A
f
(b) The assertion on ò  can be proved using supremum arguments similar to that in (a) by
A
noting that for  > 0 and jÎ B (E), j/ £ f on A whenever j £ f on A, and j£ f on A
0
whenever j £ f on A.
To verify (f g)+ = f + A ò g , note that if j, j Î B (E) and j £ f, j £ g on A, then j + j Î B (E)
A ò
A ò
0
0
and j + j £ f + g on A so
+
+
A ò (f g)  ò A (j + j ) (by definition of ò A (f g))
= ò j + A ò j
A
take supremum over all such j’s and j ’s we have (f g)+  A ò f + A ò g. For the opposite inequality,
A ò
note that if Î B (E) with £ f + g on A, then write j = min {, f} and j =  – j we see that j, j
0
Î B (E) (note (i) – M £ j £  £ M if || £ M so j is bounded on E; (ii) j =  – j is bounded on E
0
because both  and j are; (iii) measurability of j, j is clear; and (iv) from j = min {, f} and j
= max {0,  – f} we see that j, j = 0 whenever  = 0 so j, j vanishes outside a set of finite
measure). Further, we have j £ f, j £ g on A. Hence

A ò  = ò j + j
A ò
A
£ ò A f A ò + g
+
+
Taking supremum over all such ’s we get ò A (f g) £ A ò f g
Theorem 1: Fatou’s Lemma
Suppose {f } is a sequence of non-negative measurable functions defined on E and {f } converges
n n
(pointwisely) to a non-negative function f a.e. on E. Then

E ò f £ lim inf f n
E ò
n
Proof: Let h Î B (E) and h £ f on E. Then there exists A  E with m(A) <  such that h = 0 outside
0
A. Let h = min {f , h} on A, we have h is uniformly bounded and measurable on A : in fact if
n n n
|h| £ M on E, then h = min {f , h} > min {0, h}  –M and h = min {f , h} £ h £ M so |h | £ M on
n n n n n
A Further, with the observation that min {a, b} = (a + b – |a – b|)/2 for all real a, b we have
+
-
-
f + h |f - h| f h |f h|
-
h = n n  = min {f, h} = h
n
2 2
on A. Since m(A) < , we can conclude by Bounded Convergence Theorem that ò h = lim A ò h n .
A
n
So assuming h = 0 on E\A, we have
n
E ò
ò
E ò h = A ò h lim A ò h = lim h £ liminf f n
n
E n
n n n
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