Monotone Convergence Theorem and its Conditions
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I am reading Folland's Real Analysis p. 51
Corollary:
If ${f_n} subset L^+$, $fin L^+$, and $f_n(x)$ increases to $f(x)$ for a.e. $x$, then $int f =lim int f_n $.
This is a bit relaxed from monotone convergence theorem since here we require only "a.e. x".
Then, the author gives two examples: $$chi_{(n,n+1)} rightarrow 0 text{and} n_{chi(0,1/n)}rightarrow 0 textbf{pointwise},$$
but $$int chi_{(n,n+1)} = int n_{chi(0,1/n)} = 1,$$
so obvious, both of them do not meet MCT (otherwise, both integration should be $0$). I cannot understand the explanation from the author.
Is this because of the difference between "pointwise" and "almost everywhere"?
real-analysis measure-theory lebesgue-integral
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
$endgroup$
add a comment |
$begingroup$
I am reading Folland's Real Analysis p. 51
Corollary:
If ${f_n} subset L^+$, $fin L^+$, and $f_n(x)$ increases to $f(x)$ for a.e. $x$, then $int f =lim int f_n $.
This is a bit relaxed from monotone convergence theorem since here we require only "a.e. x".
Then, the author gives two examples: $$chi_{(n,n+1)} rightarrow 0 text{and} n_{chi(0,1/n)}rightarrow 0 textbf{pointwise},$$
but $$int chi_{(n,n+1)} = int n_{chi(0,1/n)} = 1,$$
so obvious, both of them do not meet MCT (otherwise, both integration should be $0$). I cannot understand the explanation from the author.
Is this because of the difference between "pointwise" and "almost everywhere"?
real-analysis measure-theory lebesgue-integral
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
$endgroup$
6
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no... it's because of "increases"
$endgroup$
– mathworker21
Dec 27 '18 at 8:03
add a comment |
$begingroup$
I am reading Folland's Real Analysis p. 51
Corollary:
If ${f_n} subset L^+$, $fin L^+$, and $f_n(x)$ increases to $f(x)$ for a.e. $x$, then $int f =lim int f_n $.
This is a bit relaxed from monotone convergence theorem since here we require only "a.e. x".
Then, the author gives two examples: $$chi_{(n,n+1)} rightarrow 0 text{and} n_{chi(0,1/n)}rightarrow 0 textbf{pointwise},$$
but $$int chi_{(n,n+1)} = int n_{chi(0,1/n)} = 1,$$
so obvious, both of them do not meet MCT (otherwise, both integration should be $0$). I cannot understand the explanation from the author.
Is this because of the difference between "pointwise" and "almost everywhere"?
real-analysis measure-theory lebesgue-integral
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
$endgroup$
I am reading Folland's Real Analysis p. 51
Corollary:
If ${f_n} subset L^+$, $fin L^+$, and $f_n(x)$ increases to $f(x)$ for a.e. $x$, then $int f =lim int f_n $.
This is a bit relaxed from monotone convergence theorem since here we require only "a.e. x".
Then, the author gives two examples: $$chi_{(n,n+1)} rightarrow 0 text{and} n_{chi(0,1/n)}rightarrow 0 textbf{pointwise},$$
but $$int chi_{(n,n+1)} = int n_{chi(0,1/n)} = 1,$$
so obvious, both of them do not meet MCT (otherwise, both integration should be $0$). I cannot understand the explanation from the author.
Is this because of the difference between "pointwise" and "almost everywhere"?
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
asked Dec 27 '18 at 7:53
sleeve chensleeve chen
3,13041852
3,13041852
6
$begingroup$
no... it's because of "increases"
$endgroup$
– mathworker21
Dec 27 '18 at 8:03
add a comment |
6
$begingroup$
no... it's because of "increases"
$endgroup$
– mathworker21
Dec 27 '18 at 8:03
6
6
$begingroup$
no... it's because of "increases"
$endgroup$
– mathworker21
Dec 27 '18 at 8:03
$begingroup$
no... it's because of "increases"
$endgroup$
– mathworker21
Dec 27 '18 at 8:03
add a comment |
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$begingroup$
no... it's because of "increases"
$endgroup$
– mathworker21
Dec 27 '18 at 8:03