Monotone Convergence Theorem and its Conditions












0














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"?










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  • 6




    no... it's because of "increases"
    – mathworker21
    Dec 27 '18 at 8:03
















0














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"?










share|cite|improve this question


















  • 6




    no... it's because of "increases"
    – mathworker21
    Dec 27 '18 at 8:03














0












0








0







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"?










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 7:53









sleeve chen

3,05941852




3,05941852








  • 6




    no... it's because of "increases"
    – mathworker21
    Dec 27 '18 at 8:03














  • 6




    no... it's because of "increases"
    – mathworker21
    Dec 27 '18 at 8:03








6




6




no... it's because of "increases"
– mathworker21
Dec 27 '18 at 8:03




no... it's because of "increases"
– mathworker21
Dec 27 '18 at 8:03










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