Page 122 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 10: Measurable Functions

Notes
n
Similarly fg = a k b k E k
k 1
which is again a linear combination of characteristic function, therefore fg is simple.
Theorem 7: Let E be a measurable set with m (E) < and {f } a sequence of measurable functions
n
converging a.e. to a real valued function defined on E. Then, given > 0 and > 0, there is a set
A E with m (A) < and an integer N such that |f (x) – f (x)| < for all x E – A and all n N.
n
Proof: Let F be the set of points of E for which f f. Then m (F) = 0 and f (x) f (x) for all x
n n
E – F = E (say). Then by the previous theorem for the set E , we get A E with m (A ) < and
1 1 1 1 1
an integer N such that
|f (x) – f (x) | < for all n N and x E – A .
n 1 1
We get the required result by taking
A = A F since m (F) = 0 and E – A = E – A
1 1 1

Note Before proving this theorem first prove the previous theorem.

10.1.8 Convergent Sequence of Measurable Function

Definition: A sequence {f } of measurable functions is said to converge almost uniformly to a
n
measurable function f defined on a measurable set E if for each > 0 there exists a measurable
set A E with m (A) < such that {f } converges to f uniformly an E – A.
n
10.1.9 Egoroff's Theorem

Statement: Let E be a measurable set with m (E) < and {f } a sequence of measurable functions
n
which converge to f a.e. on E. Then, given > 0 there is a set A E with m (A) < with that the
sequence {f } converges to f uniformly on E – A.
n
Proof: Applying the theorem, “Let E be a measurable set with m (E) < and {f } a sequence of
n
measurable function converging a.e. to real valued function f defined on E. Then given > 0 and
> 0 there is a set A E with m (A) < and an integer N such that
|f (x) – f (x) | < for all x E – A and all n N”
n
with = 1, = /2, we get a measurable set
A E with m (A ) < /2 and a positive integer N , such that
1 1 1
|f (x) – f (x)|< 1 for all x N
n 1
and x E , where E = E – A .
1 1 1
Again, taking = 1/2 and = n/2 ,
2
2
we get a measurable set A E with m (A ) < /2 , and a positive integer N such that
2 1 2 2
1
|f (x) – f (x)| < n N and x E where E = E – A , and so on.
n 2 2 2 1 2
2

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