Page 136 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 11: Integration

Notes
(f g)
1 2
E E E

Since inf 1 = f
f 1
E E
and inf = g
g 2 2
E E

(f g) f g … (2)
E E E
On the other hand if and are two simple functions such that < f and g. Then + is
1 2 1 2 1 2
simple function and
+ f + g,
1 2
or f + g +
1 2

(f g) ( 1 2 )
E E

But ( ) =
1 2 1 2
E E E
(f g) 1 2
E E E

Since sup = f
1
f 1
E E
and sup 2 = g
f 2
E E
(f g) f g … (3)
E E E
From (2) and (3), we get

(f g) = f g
E E E

(af bg) = af bg
E E E

= a f b g from (i)
E E
Proof of 2: Since f = g a.e.
f – g = 0 a.e.

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