Page 383 - DMTH401_REAL ANALYSIS

Unit 32: The General Lebesgue Integral and Convergence in Measure

Theorem 1: Let f and g are integrable over E. Then Notes
(i) The function cf is integrable over E, and cf = c f.
E ò
E ò
(ii) The function f + g is integrable over E, and f g+ = E ò f + E ò g .
E ò
(iii) If f g a.e., then f ò E g
E ò
(iv) If A and B are disjoint measurable sets contained in E, then ò f = A ò f + B ò f
È
A B
Proof:
(i) Since f is integrable over E, both f and f are integrable over E and the integral of f is given
+
–
by
+
E ò f = E ò f - E ò f -
Hence,
–
+
–
both cf and cf are integrable over E, and hence, cf = cf – cf are integrable over E and
+
+
E ò cf = E ò cf - E ò cf -
+
= c f - c f -
E ò
E ò
-
+
= c[ f - E ò f ]
E ò
= c f.
E ò
Hence (i) is proved.
(ii) Suppose if f and f are nonnegative integrable functions with f = f – f ,
1 2 1 2
Then f – f = f – f .
+
–
1 2
Hence,
+
f + f = f + f .
–
2 1
As you know
f + f = f – f .
+
–
2 1
Therefore,
+
f = f – f –
= f – f .
1 2
Since f and g are measurable,
–
+
f , f , g , g are measurable.
+
–
Hence,
f + g , f + g are also measurable.
+
–
+
–
–
And f + g = (f + g ) – (f + g ).
+
+
–
Hence by(1),
–
–
+
+
(f + g ) = (f + g ) – (f + g )
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