Almost sure and order convergence are equivalent in $L_p$ spaces
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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
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
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add a comment |
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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
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
$endgroup$
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Can you recall the definition of order convergence?
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– Davide Giraudo
Dec 14 '18 at 13:18
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@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 $inf_n g_n = 0$.
$endgroup$
– grndl
Dec 14 '18 at 14:14
add a comment |
$begingroup$
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
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
$endgroup$
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
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
edited Dec 14 '18 at 14:23
grndl
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
asked Dec 14 '18 at 1:25
grndlgrndl
4,53331338
4,53331338
$begingroup$
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 $inf_n g_n = 0$.
$endgroup$
– grndl
Dec 14 '18 at 14:14
add a comment |
$begingroup$
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 $inf_n g_n = 0$.
$endgroup$
– grndl
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 $inf_n g_n = 0$.
$endgroup$
– grndl
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 $inf_n g_n = 0$.
$endgroup$
– grndl
Dec 14 '18 at 14:14
add a comment |
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 $g_n := sup_{m geq n}|f_m - f|$.
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
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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 $g_n := sup_{m geq n}|f_m - f|$.
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
$begingroup$
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 $g_n := sup_{m geq n}|f_m - f|$.
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
$begingroup$
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 $g_n := sup_{m geq n}|f_m - f|$.
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
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 $g_n := sup_{m geq n}|f_m - f|$.
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
answered Dec 14 '18 at 14:22
Davide GiraudoDavide Giraudo
127k16151265
127k16151265
add a comment |
add a comment |
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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 $inf_n g_n = 0$.
$endgroup$
– grndl
Dec 14 '18 at 14:14