Proving DCT from Fatou's Lemma












1












$begingroup$


Forgive me, I am new to measure theory.
I am trying to prove the Dominated Convergence Theorem by assuming Fatou's Lemma, here is what I have so far.



Fatou's Lemma:



Let ${f_n}
$ be a sequence of nonnegative measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



then $int f leq $ lim$_{n rightarrow infty}$ inf $int f_n$.



The DCT states:



Let ${f_n}$ be sequence of measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



if in addition we have



$vert f(x) vert leq g(x)$ ; $g(x)$ integrable,



then $int vert f_n - f vert rightarrow 0$, as $n rightarrow infty$.



I know one way to prove this is you can define a set of elements bounded above by integer values
so the functions are supported on a set of finite measure allowing the use of the bounded convergence theorem.



Now Fatou's Lemma takes into consideration the nonnegative functions, something I cannot assume with the DCT, but since the $f_n$ are all bounded above by an integrable function $g(x)$ could I rewrite $g(x)$ as its decomposition into $g^+ - g^-$?



My intuition says the result will "pop out" if I had non negativity? OR am I missing something else here?






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









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  • 1




    $begingroup$
    The condition $|f| leq g$ a.e. implies $g geq 0$ a.e.
    $endgroup$
    – Will M.
    Dec 19 '18 at 23:22










  • $begingroup$
    omg I cannot believe I missed that, thanks!!
    $endgroup$
    – Hossien Sahebjame
    Dec 22 '18 at 20:53
















1












$begingroup$


Forgive me, I am new to measure theory.
I am trying to prove the Dominated Convergence Theorem by assuming Fatou's Lemma, here is what I have so far.



Fatou's Lemma:



Let ${f_n}
$ be a sequence of nonnegative measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



then $int f leq $ lim$_{n rightarrow infty}$ inf $int f_n$.



The DCT states:



Let ${f_n}$ be sequence of measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



if in addition we have



$vert f(x) vert leq g(x)$ ; $g(x)$ integrable,



then $int vert f_n - f vert rightarrow 0$, as $n rightarrow infty$.



I know one way to prove this is you can define a set of elements bounded above by integer values
so the functions are supported on a set of finite measure allowing the use of the bounded convergence theorem.



Now Fatou's Lemma takes into consideration the nonnegative functions, something I cannot assume with the DCT, but since the $f_n$ are all bounded above by an integrable function $g(x)$ could I rewrite $g(x)$ as its decomposition into $g^+ - g^-$?



My intuition says the result will "pop out" if I had non negativity? OR am I missing something else here?






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The condition $|f| leq g$ a.e. implies $g geq 0$ a.e.
    $endgroup$
    – Will M.
    Dec 19 '18 at 23:22










  • $begingroup$
    omg I cannot believe I missed that, thanks!!
    $endgroup$
    – Hossien Sahebjame
    Dec 22 '18 at 20:53














1












1








1


1



$begingroup$


Forgive me, I am new to measure theory.
I am trying to prove the Dominated Convergence Theorem by assuming Fatou's Lemma, here is what I have so far.



Fatou's Lemma:



Let ${f_n}
$ be a sequence of nonnegative measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



then $int f leq $ lim$_{n rightarrow infty}$ inf $int f_n$.



The DCT states:



Let ${f_n}$ be sequence of measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



if in addition we have



$vert f(x) vert leq g(x)$ ; $g(x)$ integrable,



then $int vert f_n - f vert rightarrow 0$, as $n rightarrow infty$.



I know one way to prove this is you can define a set of elements bounded above by integer values
so the functions are supported on a set of finite measure allowing the use of the bounded convergence theorem.



Now Fatou's Lemma takes into consideration the nonnegative functions, something I cannot assume with the DCT, but since the $f_n$ are all bounded above by an integrable function $g(x)$ could I rewrite $g(x)$ as its decomposition into $g^+ - g^-$?



My intuition says the result will "pop out" if I had non negativity? OR am I missing something else here?






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$




Forgive me, I am new to measure theory.
I am trying to prove the Dominated Convergence Theorem by assuming Fatou's Lemma, here is what I have so far.



Fatou's Lemma:



Let ${f_n}
$ be a sequence of nonnegative measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



then $int f leq $ lim$_{n rightarrow infty}$ inf $int f_n$.



The DCT states:



Let ${f_n}$ be sequence of measurable functions s.t.



$f_n rightarrow f$ a.e. x as $n rightarrow infty$,



if in addition we have



$vert f(x) vert leq g(x)$ ; $g(x)$ integrable,



then $int vert f_n - f vert rightarrow 0$, as $n rightarrow infty$.



I know one way to prove this is you can define a set of elements bounded above by integer values
so the functions are supported on a set of finite measure allowing the use of the bounded convergence theorem.



Now Fatou's Lemma takes into consideration the nonnegative functions, something I cannot assume with the DCT, but since the $f_n$ are all bounded above by an integrable function $g(x)$ could I rewrite $g(x)$ as its decomposition into $g^+ - g^-$?



My intuition says the result will "pop out" if I had non negativity? OR am I missing something else here?






real-analysis measure-theory lebesgue-integral lebesgue-measure




real-analysis measure-theory lebesgue-integral lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 22:34









Hossien SahebjameHossien Sahebjame

1049




1049








  • 1




    $begingroup$
    The condition $|f| leq g$ a.e. implies $g geq 0$ a.e.
    $endgroup$
    – Will M.
    Dec 19 '18 at 23:22










  • $begingroup$
    omg I cannot believe I missed that, thanks!!
    $endgroup$
    – Hossien Sahebjame
    Dec 22 '18 at 20:53














  • 1




    $begingroup$
    The condition $|f| leq g$ a.e. implies $g geq 0$ a.e.
    $endgroup$
    – Will M.
    Dec 19 '18 at 23:22










  • $begingroup$
    omg I cannot believe I missed that, thanks!!
    $endgroup$
    – Hossien Sahebjame
    Dec 22 '18 at 20:53








1




1




$begingroup$
The condition $|f| leq g$ a.e. implies $g geq 0$ a.e.
$endgroup$
– Will M.
Dec 19 '18 at 23:22




$begingroup$
The condition $|f| leq g$ a.e. implies $g geq 0$ a.e.
$endgroup$
– Will M.
Dec 19 '18 at 23:22












$begingroup$
omg I cannot believe I missed that, thanks!!
$endgroup$
– Hossien Sahebjame
Dec 22 '18 at 20:53




$begingroup$
omg I cannot believe I missed that, thanks!!
$endgroup$
– Hossien Sahebjame
Dec 22 '18 at 20:53










1 Answer
1






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oldest

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$begingroup$

$int [g-f]=int lim inf [g-f_n] leq lim inf int [g-f_n]$ which gives $int g -int f leq int g -lim sup int f_n$ so $lim sup int f_n leq int f$. Now replace $f$ by $-f$ and $f_n$ by $-f_n$ to get $lim inf int f_n geq int f$. Hence $int f_n to int f$. To get $int |f_n-f| to 0$ you simply have to replace $f$ by $0$, $g$ by $2g$ and $f_n$ by $|f_n-f|$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok I finally understand it haha thanks so much man!!!
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 3:17











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4












$begingroup$

$int [g-f]=int lim inf [g-f_n] leq lim inf int [g-f_n]$ which gives $int g -int f leq int g -lim sup int f_n$ so $lim sup int f_n leq int f$. Now replace $f$ by $-f$ and $f_n$ by $-f_n$ to get $lim inf int f_n geq int f$. Hence $int f_n to int f$. To get $int |f_n-f| to 0$ you simply have to replace $f$ by $0$, $g$ by $2g$ and $f_n$ by $|f_n-f|$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok I finally understand it haha thanks so much man!!!
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 3:17
















4












$begingroup$

$int [g-f]=int lim inf [g-f_n] leq lim inf int [g-f_n]$ which gives $int g -int f leq int g -lim sup int f_n$ so $lim sup int f_n leq int f$. Now replace $f$ by $-f$ and $f_n$ by $-f_n$ to get $lim inf int f_n geq int f$. Hence $int f_n to int f$. To get $int |f_n-f| to 0$ you simply have to replace $f$ by $0$, $g$ by $2g$ and $f_n$ by $|f_n-f|$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok I finally understand it haha thanks so much man!!!
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 3:17














4












4








4





$begingroup$

$int [g-f]=int lim inf [g-f_n] leq lim inf int [g-f_n]$ which gives $int g -int f leq int g -lim sup int f_n$ so $lim sup int f_n leq int f$. Now replace $f$ by $-f$ and $f_n$ by $-f_n$ to get $lim inf int f_n geq int f$. Hence $int f_n to int f$. To get $int |f_n-f| to 0$ you simply have to replace $f$ by $0$, $g$ by $2g$ and $f_n$ by $|f_n-f|$






share|cite|improve this answer









$endgroup$



$int [g-f]=int lim inf [g-f_n] leq lim inf int [g-f_n]$ which gives $int g -int f leq int g -lim sup int f_n$ so $lim sup int f_n leq int f$. Now replace $f$ by $-f$ and $f_n$ by $-f_n$ to get $lim inf int f_n geq int f$. Hence $int f_n to int f$. To get $int |f_n-f| to 0$ you simply have to replace $f$ by $0$, $g$ by $2g$ and $f_n$ by $|f_n-f|$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 23:16









Kavi Rama MurthyKavi Rama Murthy

62.7k42262




62.7k42262












  • $begingroup$
    ok I finally understand it haha thanks so much man!!!
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 3:17


















  • $begingroup$
    ok I finally understand it haha thanks so much man!!!
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 3:17
















$begingroup$
ok I finally understand it haha thanks so much man!!!
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 3:17




$begingroup$
ok I finally understand it haha thanks so much man!!!
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 3:17


















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