7/28/2019 Proof of Monotone Convergence Theorem
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proof of monotone convergence theorem
proof of monotone convergence
theorem
It is enough to prove the following
Theorem 1.
Let(X,) be a measurable space and let fk:X {+} be a monotone
increasing
sequence of positive measurable functions (i.e. 0 f1 f2 ).
Then
f(x) = limk fk(x) is measurable and
lim
n
X
fkd =X
f(x)d .
First of all by the monotonicity
(http://planetmath.org/node/40195) of the sequence
(http://planetmath.org/node/30397) we have
f(x) = supk
fk(x)
hence we know that f is measurable
(http://planetmath.org/node/36778) . Moreover being
fk ffor all k, by the monotonicity of the integral
(http://planetmath.org/node/31920) , we
immediately get
supk
Xfkd
Xf(x)d .
So take any simple (http://planetmath.org/node/32189) measurable
function
(http://planetmath.org/node/33176) s such that 0 s f. Given also
< 1 define
Ek= {xX: fk(x) s(x)} .
The sequence Ekis an increasing
(http://planetmath.org/node/34228) sequence of
measurable sets (http://planetmath.org/node/30755) . Moreover
the union (http://planetmath.org
/node/34394)of all Ekis the whole spaceXsince limk fk(x) = f(x)
s(x) > s(x).
Moreover it holds
X
fkd Ek
fkd Ek
sd .
Since s is a simple measurable function it is easy to check that
E Esd is a
measure (http://planetmath.org/node/40498) and hence
supk
Xfkd
Xsd .
But this last inequality (http://planetmath.org/node/30000)
holds for every < 1 and for all
simple measurable functions s with s f. Hence by the definition
(http://planetmath.org
/node/39936)ofLebesgue integral
(http://planetmath.org/node/31920)
supk
Xfkd
Xf d
which completes (http://planetmath.org/node/33445) the
proof.
Major Section:
Reference
Type of Math Object:
Proof
Parent:
monotone convergence theorem (/monotoneconvergencetheorem)
Mathematics Subject Classification
28A20 Measurable and nonmeasurable functions, sequences of
measurable functions,
modes of convergence (/msc_browser/28A20)
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f of monotone convergence theorem | planetmath.org
http://planetmath.org/proofofmonotoneconvergencetheorem
2 2013-07-17 00:02
http://planetmath.org/node/40195http://planetmath.org/node/30397http://planetmath.org/node/36778http://planetmath.org/node/31920http://planetmath.org/node/32189http://planetmath.org/node/33176http://planetmath.org/node/34228http://planetmath.org/node/30755http://planetmath.org/http://planetmath.org/node/40498http://planetmath.org/node/30000http://planetmath.org/http://planetmath.org/node/31920http://planetmath.org/node/33445http://planetmath.org/proofofmonotoneconvergencetheoremhttp://planetmath.org/proofofmonotoneconvergencetheoremhttp://planetmath.org/node/33445http://planetmath.org/node/31920http://planetmath.org/http://planetmath.org/node/30000http://planetmath.org/node/40498http://planetmath.org/http://planetmath.org/node/30755http://planetmath.org/node/34228http://planetmath.org/node/33176http://planetmath.org/node/32189http://planetmath.org/node/31920http://planetmath.org/node/36778http://planetmath.org/node/30397http://planetmath.org/node/40195
7/28/2019 Proof of Monotone Convergence Theorem
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Comments
The last inequality (/comment/2103#comment-2103)
Permalink (/comment/2103#comment-2103)Submitted by bchui
(/users/bchui) on Tue,11/14/2006 - 12:14
Should the last inequality be
$\displaystyle{ \int_X fd\mu \leq \sup_{k}\int_X f_k d\mu }$
instead of $\int_k$ on the RHS ?
Re: The last inequality (/comment/2106#comment-2106)
Permalink (/comment/2106#comment-2106)Submitted by paolini
(/users/paolini) on Wed,
11/15/2006 - 13:02
Yes, thanks. I corrected it.
E.
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