Page 116 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 10: Measurable Functions

Now E (f – g > a) = E (f > a + g) Notes
since g is measurable.
cg is measurable, c is constant.

a + cg is measurable a, c R
a + g is measurable by taking c = 1,
since f and a + g are measurable
E (f > a + g) is measurable.

E (f – g > a) is a measurable set.
f – g is a measurable function.
(e) To prove |f| is measurable.
We have

E if 0
E (|f| > ) =
[E (f )] [E(f )]if 0

[because we know that |x| > a x > a or x < –a]
since f is measurable therefore E (f > ) and E (f < – ) are measurable by definition.
Also we know that finite union of two measurable sets is measurable.
E (f > ) E (f < – ) is measurable.
E (|f| > ) is measurable.

|f| is measurable.
2
(f) To prove f is measurable.
E if 0
We have E (f > ) =
2
E (|f| )]if 0
But E (|f| > ) = [E(f )] [E(f )], if 0 ( |x| > a x > a or x < – a)

E if 0
E (f > ) =
2
[E (f )] [E(f )]if 0
But f is measurable over E.

E (f ) and E (f ) are measurable sets.

[E (f )] [E(f )] is measurable.

( union of two measurable sets is measurable)
E (f > ) is measurable because both the sets on RHS are measurable.
2
2
f is measurable over E.
(g) To prove fg is measurable.

Clearly, f + g and f – g are measurable functions over E.

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