Page 386 - DMTH401_REAL ANALYSIS

Real Analysis

Notes And f (x) = 0 otherwise.
n


Then m{x/|fn(x)| > } = 2 – 2/n [since 2 n < 2 + 1]
Hence f > 0 in measure.
n

Notes That the sequence {f (x)} has the value 1 for arbitrarily large values of n.
n
Hence {f (x)} does not converge for any x in [0, 1].
n
Theorem 4: Let {f } be a sequence of measurable functions that converges in measure to f.
n
Then there is a subsequence {f } that converges to f almost everywhere.
nk
Proof: Since {f } is a sequence of measurable functions that converges in measure to f,
n
Given , there is an integer n such that for all n n,

-
-
m{x/ f(x) f (x) 2 } < 2 - ...(1)
n
–
Let E = {x/|f (x) – f(x)| 2 }
 nv
Therefore,
¥
if x Ï  E u
u= k
–
then |f (x) – f(x)| < 2 for k.
n
Therefore,
F (x) > f(x).
n
¥ ¥
Hence f (x) > f(x) for any x AÏ   E
n u
=
k 1 u= k
¥
é
But mA m  E ù
ê ë u= k u ú û
¥
å mE u
u= k
= 2 –k+1 .

Hence mA = 0
Theorem 5: Let {f } be a sequence of measurable functions defined on a measurable set E of finite
n
measure.
Then {f } converges to f in measure if and only if every subsequence of {f } has in turn a subsequence
n n
that converges almost everywhere to f.
Theorem 6: Fatou’s lemma and the monotone and Lebesgue Convergence theorem remain valid
if ‘convergence a.e.’ is replaced by ‘convergence in measure’.

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