Page 115 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Consider the case in which c 0, then

E f if c 0
c
E (cf > ) =
E f if c 0
c
Both the sets on R.H.S. are measurable.

Hence E (cf > ) is measurable and so cf is measurable c R.
(c) Before proving f + g is measurable, we first prove that if f and g are measurable over E then
the set E (f > g) is also measurable.

Now f > g a rational number r such that
f (x) > r > g (x)


Thus E (f > g) = [(E (f r) (E(g r))]
r Q
= an enumerable union of measurable sets.
= measurable set, since Q is an enumerable set.

Now, we shall prove that f + g is measurable over E. Let a be any real number.
Now E (f + g > a) = E (f > a – g) … (1)
Again, g is measurable
cg is measurable, c is constant.

( We know that if f is a measurable function and c is constant then cf is measurable)
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.

(d) To prove that f – g is measurable. Before proving f – g is measurable, we first prove that if
f and g are measurable over E then the set E (f > g) is also measurable.
Now f > g a rational number r, such that f (x) > r > g (x).


Thus E (f > g) = [(E (f r) (E(g r))]
r Q
= an enumerable union of measurable sets.

= measurable sets, since Q is an enumerable set.
Now we shall prove that f – g is measurable over E.
Let a be any real number.

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