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Unit 10: Measurable Functions
Notes
Note Egoroff’s theorem can be stated as: almost every where convergence implies almost
uniform convergence.
10.1.10 Riesz Theorem
Let {f } be a sequence of measurable functions which converges in measure to f. Then there is a
n
subsequence f which converges in measure to f a.e.
n k
Proof: Let { } and { } be two sequences of positive real numbers such that 0 as n and
n n n
.
n
n 1
Let us now choose an increasing sequence {n } of positive integers as follows.
k
Let n be a positive integer such that
m x E f (x) f(x)
n
Since f f in measure for a given > 0 and > 0, a positive integer n such that
n 1 1 1
m x E, f (x) f(x) , n n
n 1 1 1 1
Similarly, let n be a positive number such that m x E, f (x) f(x) , n n and so
2 n 2 2 2 2 1
on.
In general let n be a positive number such that
k
m x : x E, f (x) f(x) and that n n .
n k k k k k 1
We shall now prove that the subsequence f converges to f a.e.
n k
k
Let A = x : x E, f (x) f(x) i ,k N and A A .
k
n i
i k k 1
Clearly, {A } is a decreasing sequence of measurable sets and m (A ) < .
k 1
Therefore, we have
m (A) = limm(A )
k
k
But m (A ) 0 as k .
k i
i k
Hence m (A) = 0.
Now it remains to show that {f } converges to f on E – A. Let x E – A.
n o
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