Almost sure and order convergence are equivalent in $L_p$ spaces












1












$begingroup$


I have attempted to prove the following lemma, which is given as an exercise by Aliprantis and Border. Is the proof correct? Improvements are welcome.



Lemma. An order bounded sequence $(f_n)$ in some $L_p(mu)$ space satisfies $f_n xrightarrow{o} f$ if and only if $f_n to f$ a.s. ($mu$). (The symbol $xrightarrow{o}$ denotes order convergence.)



Proof. First note that since $(f_n)$ is order bounded $|f_n - f| leq h + |f|$ holds for some $h in L_p(mu)$ and all $n$. Therefore, $sup_{m geq n}|f_n - f| in L_p(mu)$ for all $n$.



Assume $f_n xrightarrow{o} f$, which means that there exists a sequence $(g_n) in L_p(mu)$ such that $|f_n - f| leq g_n downarrow 0$. The inequality means that $|f_n(x) - f(x)| leq g_n(x)$ for $mu$-almost every $x$ and all $n$. Thus, $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$ for $mu$-almost every $x$. This implies $f_n to f$ a.s. ($mu$).



Now assume $f_n to f$ a.s. ($mu$). Let $g_n := sup_{m geq n}|f_m - f|$. Then, $|f_n - f| leq g_n downarrow 0$.






measure-theory proof-verification convergence lp-spaces vector-lattices







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  • $begingroup$
    Can you recall the definition of order convergence?
    $endgroup$
    – Davide Giraudo
    Dec 14 '18 at 13:18










  • $begingroup$
    @DavideGiraudo If $V$ is a vector lattice and $(f_n), f in V$, then $f_n xrightarrow{o} f$ if and only if there exists a decreasing sequence $(g_n) in V$ such that $|f_n - f| leq g_n$ for all $n$ and $inf_n g_n = 0$.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:14
















1












$begingroup$


I have attempted to prove the following lemma, which is given as an exercise by Aliprantis and Border. Is the proof correct? Improvements are welcome.



Lemma. An order bounded sequence $(f_n)$ in some $L_p(mu)$ space satisfies $f_n xrightarrow{o} f$ if and only if $f_n to f$ a.s. ($mu$). (The symbol $xrightarrow{o}$ denotes order convergence.)



Proof. First note that since $(f_n)$ is order bounded $|f_n - f| leq h + |f|$ holds for some $h in L_p(mu)$ and all $n$. Therefore, $sup_{m geq n}|f_n - f| in L_p(mu)$ for all $n$.



Assume $f_n xrightarrow{o} f$, which means that there exists a sequence $(g_n) in L_p(mu)$ such that $|f_n - f| leq g_n downarrow 0$. The inequality means that $|f_n(x) - f(x)| leq g_n(x)$ for $mu$-almost every $x$ and all $n$. Thus, $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$ for $mu$-almost every $x$. This implies $f_n to f$ a.s. ($mu$).



Now assume $f_n to f$ a.s. ($mu$). Let $g_n := sup_{m geq n}|f_m - f|$. Then, $|f_n - f| leq g_n downarrow 0$.






measure-theory proof-verification convergence lp-spaces vector-lattices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you recall the definition of order convergence?
    $endgroup$
    – Davide Giraudo
    Dec 14 '18 at 13:18










  • $begingroup$
    @DavideGiraudo If $V$ is a vector lattice and $(f_n), f in V$, then $f_n xrightarrow{o} f$ if and only if there exists a decreasing sequence $(g_n) in V$ such that $|f_n - f| leq g_n$ for all $n$ and $inf_n g_n = 0$.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:14














1












1








1





$begingroup$


I have attempted to prove the following lemma, which is given as an exercise by Aliprantis and Border. Is the proof correct? Improvements are welcome.



Lemma. An order bounded sequence $(f_n)$ in some $L_p(mu)$ space satisfies $f_n xrightarrow{o} f$ if and only if $f_n to f$ a.s. ($mu$). (The symbol $xrightarrow{o}$ denotes order convergence.)



Proof. First note that since $(f_n)$ is order bounded $|f_n - f| leq h + |f|$ holds for some $h in L_p(mu)$ and all $n$. Therefore, $sup_{m geq n}|f_n - f| in L_p(mu)$ for all $n$.



Assume $f_n xrightarrow{o} f$, which means that there exists a sequence $(g_n) in L_p(mu)$ such that $|f_n - f| leq g_n downarrow 0$. The inequality means that $|f_n(x) - f(x)| leq g_n(x)$ for $mu$-almost every $x$ and all $n$. Thus, $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$ for $mu$-almost every $x$. This implies $f_n to f$ a.s. ($mu$).



Now assume $f_n to f$ a.s. ($mu$). Let $g_n := sup_{m geq n}|f_m - f|$. Then, $|f_n - f| leq g_n downarrow 0$.






measure-theory proof-verification convergence lp-spaces vector-lattices







share|cite|improve this question











$endgroup$




I have attempted to prove the following lemma, which is given as an exercise by Aliprantis and Border. Is the proof correct? Improvements are welcome.



Lemma. An order bounded sequence $(f_n)$ in some $L_p(mu)$ space satisfies $f_n xrightarrow{o} f$ if and only if $f_n to f$ a.s. ($mu$). (The symbol $xrightarrow{o}$ denotes order convergence.)



Proof. First note that since $(f_n)$ is order bounded $|f_n - f| leq h + |f|$ holds for some $h in L_p(mu)$ and all $n$. Therefore, $sup_{m geq n}|f_n - f| in L_p(mu)$ for all $n$.



Assume $f_n xrightarrow{o} f$, which means that there exists a sequence $(g_n) in L_p(mu)$ such that $|f_n - f| leq g_n downarrow 0$. The inequality means that $|f_n(x) - f(x)| leq g_n(x)$ for $mu$-almost every $x$ and all $n$. Thus, $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$ for $mu$-almost every $x$. This implies $f_n to f$ a.s. ($mu$).



Now assume $f_n to f$ a.s. ($mu$). Let $g_n := sup_{m geq n}|f_m - f|$. Then, $|f_n - f| leq g_n downarrow 0$.






measure-theory proof-verification convergence lp-spaces vector-lattices




measure-theory proof-verification convergence lp-spaces vector-lattices






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edited Dec 14 '18 at 14:23







aduh

















asked Dec 14 '18 at 1:25









aduhaduh

4,48631338




4,48631338












  • $begingroup$
    Can you recall the definition of order convergence?
    $endgroup$
    – Davide Giraudo
    Dec 14 '18 at 13:18










  • $begingroup$
    @DavideGiraudo If $V$ is a vector lattice and $(f_n), f in V$, then $f_n xrightarrow{o} f$ if and only if there exists a decreasing sequence $(g_n) in V$ such that $|f_n - f| leq g_n$ for all $n$ and $inf_n g_n = 0$.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:14


















  • $begingroup$
    Can you recall the definition of order convergence?
    $endgroup$
    – Davide Giraudo
    Dec 14 '18 at 13:18










  • $begingroup$
    @DavideGiraudo If $V$ is a vector lattice and $(f_n), f in V$, then $f_n xrightarrow{o} f$ if and only if there exists a decreasing sequence $(g_n) in V$ such that $|f_n - f| leq g_n$ for all $n$ and $inf_n g_n = 0$.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:14
















$begingroup$
Can you recall the definition of order convergence?
$endgroup$
– Davide Giraudo
Dec 14 '18 at 13:18




$begingroup$
Can you recall the definition of order convergence?
$endgroup$
– Davide Giraudo
Dec 14 '18 at 13:18












$begingroup$
@DavideGiraudo If $V$ is a vector lattice and $(f_n), f in V$, then $f_n xrightarrow{o} f$ if and only if there exists a decreasing sequence $(g_n) in V$ such that $|f_n - f| leq g_n$ for all $n$ and $inf_n g_n = 0$.
$endgroup$
– aduh
Dec 14 '18 at 14:14




$begingroup$
@DavideGiraudo If $V$ is a vector lattice and $(f_n), f in V$, then $f_n xrightarrow{o} f$ if and only if there exists a decreasing sequence $(g_n) in V$ such that $|f_n - f| leq g_n$ for all $n$ and $inf_n g_n = 0$.
$endgroup$
– aduh
Dec 14 '18 at 14:14










1 Answer
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The proof looks correct, up to two small typos:





  • $limsup_n |f_m(x) - f(x)| leq limsup_n g_n(x)=0$ should be $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$;

  • in the last line, the definition of $g_n$ should be $g_n := sup_{m geq n}|f_m - f|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I fixed the typos. I will leave this unanswered for a little bit just in case anyone else wants to weigh in.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:25











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1 Answer
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$begingroup$

The proof looks correct, up to two small typos:





  • $limsup_n |f_m(x) - f(x)| leq limsup_n g_n(x)=0$ should be $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$;

  • in the last line, the definition of $g_n$ should be $g_n := sup_{m geq n}|f_m - f|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I fixed the typos. I will leave this unanswered for a little bit just in case anyone else wants to weigh in.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:25
















1












$begingroup$

The proof looks correct, up to two small typos:





  • $limsup_n |f_m(x) - f(x)| leq limsup_n g_n(x)=0$ should be $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$;

  • in the last line, the definition of $g_n$ should be $g_n := sup_{m geq n}|f_m - f|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I fixed the typos. I will leave this unanswered for a little bit just in case anyone else wants to weigh in.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:25














1












1








1





$begingroup$

The proof looks correct, up to two small typos:





  • $limsup_n |f_m(x) - f(x)| leq limsup_n g_n(x)=0$ should be $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$;

  • in the last line, the definition of $g_n$ should be $g_n := sup_{m geq n}|f_m - f|$.






share|cite|improve this answer









$endgroup$



The proof looks correct, up to two small typos:





  • $limsup_n |f_m(x) - f(x)| leq limsup_n g_n(x)=0$ should be $limsup_n |f_n(x) - f(x)| leq limsup_n g_n(x)=0$;

  • in the last line, the definition of $g_n$ should be $g_n := sup_{m geq n}|f_m - f|$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 14:22









Davide GiraudoDavide Giraudo

126k16150261




126k16150261












  • $begingroup$
    Thanks! I fixed the typos. I will leave this unanswered for a little bit just in case anyone else wants to weigh in.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:25


















  • $begingroup$
    Thanks! I fixed the typos. I will leave this unanswered for a little bit just in case anyone else wants to weigh in.
    $endgroup$
    – aduh
    Dec 14 '18 at 14:25
















$begingroup$
Thanks! I fixed the typos. I will leave this unanswered for a little bit just in case anyone else wants to weigh in.
$endgroup$
– aduh
Dec 14 '18 at 14:25




$begingroup$
Thanks! I fixed the typos. I will leave this unanswered for a little bit just in case anyone else wants to weigh in.
$endgroup$
– aduh
Dec 14 '18 at 14:25


















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