My possibly fraudulent proof of $f_n$ Cauchy in measure => $f_n -> liminf$ in measure












0














This result seems too convenient and I feel like Folland would have used this to prove proposition 2.30 if this was true.



My "proof":



Let $f_n$ be Cauchy in measure.



By definition of $liminf f_n(x)$,
for every $n in mathbb{N}, x in X$ there exists an $m_{x,n}> n$ such that
$|liminf f_n(x) - f_m(x)| < epsilon / 2$



Now consider the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$}



By triangle inequality, $epsilon leq |f_n(x) -liminf f_n(x)| leq |f_n(x) - f_{m_{x,n}}(x)| + |f_{m_{x,n}}(x) - liminf f_n(x)|$



but since $|liminf f_n(x) - f_{m_{x,n}}(x)| leq epsilon / 2$, we must have that $ epsilon/2 leq |f_n(x) -liminf f_n(x)| $



therefore, the set



{$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} $subset$ {$xin X : |f_n(x) - f_{m_{x,n}}(x)| > epsilon/2$}



but the measure of the set on the right approaches to zero as $n$ tends to infinity (since $m_{x,n}> n$)



therefore the measure of the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} tends to infinity as $n$ tends to infinity



Is there anything wrong with this proof?



thank you






real-analysis measure-theory proof-verification convergence sequence-of-function







share|cite|improve this question
























  • It's liminf of what?
    – Bernard
    Dec 2 '18 at 23:24










  • sorry, lim inf fn(x). fixed.
    – Ecotistician
    Dec 2 '18 at 23:25












  • Use liminf (with a backslash) for the limit inferior.
    – Bernard
    Dec 2 '18 at 23:26






  • 1




    Yes, the proof is wrong. You only know that measure of ${x:|f_n(x)-f_m(x)| >epsilon} to 0$ as $n, m to infty$ and you cannot replace $m$ by an integer that depends on $x$.
    – Kavi Rama Murthy
    Dec 2 '18 at 23:37










  • If $liminf f_n(x)=infty$ or $-infty$ then you cannot achieve $epsilon$-closeness to $f_m(x)$ like you want to. Now, it actually happens to be true that $liminf f_n(x)in(-infty,infty)$ almost everywhere. But you would need to show that to be true. There may be other problems with your proof that I have not noticed yet.
    – Ben W
    Dec 2 '18 at 23:37
















0














This result seems too convenient and I feel like Folland would have used this to prove proposition 2.30 if this was true.



My "proof":



Let $f_n$ be Cauchy in measure.



By definition of $liminf f_n(x)$,
for every $n in mathbb{N}, x in X$ there exists an $m_{x,n}> n$ such that
$|liminf f_n(x) - f_m(x)| < epsilon / 2$



Now consider the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$}



By triangle inequality, $epsilon leq |f_n(x) -liminf f_n(x)| leq |f_n(x) - f_{m_{x,n}}(x)| + |f_{m_{x,n}}(x) - liminf f_n(x)|$



but since $|liminf f_n(x) - f_{m_{x,n}}(x)| leq epsilon / 2$, we must have that $ epsilon/2 leq |f_n(x) -liminf f_n(x)| $



therefore, the set



{$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} $subset$ {$xin X : |f_n(x) - f_{m_{x,n}}(x)| > epsilon/2$}



but the measure of the set on the right approaches to zero as $n$ tends to infinity (since $m_{x,n}> n$)



therefore the measure of the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} tends to infinity as $n$ tends to infinity



Is there anything wrong with this proof?



thank you






real-analysis measure-theory proof-verification convergence sequence-of-function







share|cite|improve this question
























  • It's liminf of what?
    – Bernard
    Dec 2 '18 at 23:24










  • sorry, lim inf fn(x). fixed.
    – Ecotistician
    Dec 2 '18 at 23:25












  • Use liminf (with a backslash) for the limit inferior.
    – Bernard
    Dec 2 '18 at 23:26






  • 1




    Yes, the proof is wrong. You only know that measure of ${x:|f_n(x)-f_m(x)| >epsilon} to 0$ as $n, m to infty$ and you cannot replace $m$ by an integer that depends on $x$.
    – Kavi Rama Murthy
    Dec 2 '18 at 23:37










  • If $liminf f_n(x)=infty$ or $-infty$ then you cannot achieve $epsilon$-closeness to $f_m(x)$ like you want to. Now, it actually happens to be true that $liminf f_n(x)in(-infty,infty)$ almost everywhere. But you would need to show that to be true. There may be other problems with your proof that I have not noticed yet.
    – Ben W
    Dec 2 '18 at 23:37














0












0








0







This result seems too convenient and I feel like Folland would have used this to prove proposition 2.30 if this was true.



My "proof":



Let $f_n$ be Cauchy in measure.



By definition of $liminf f_n(x)$,
for every $n in mathbb{N}, x in X$ there exists an $m_{x,n}> n$ such that
$|liminf f_n(x) - f_m(x)| < epsilon / 2$



Now consider the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$}



By triangle inequality, $epsilon leq |f_n(x) -liminf f_n(x)| leq |f_n(x) - f_{m_{x,n}}(x)| + |f_{m_{x,n}}(x) - liminf f_n(x)|$



but since $|liminf f_n(x) - f_{m_{x,n}}(x)| leq epsilon / 2$, we must have that $ epsilon/2 leq |f_n(x) -liminf f_n(x)| $



therefore, the set



{$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} $subset$ {$xin X : |f_n(x) - f_{m_{x,n}}(x)| > epsilon/2$}



but the measure of the set on the right approaches to zero as $n$ tends to infinity (since $m_{x,n}> n$)



therefore the measure of the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} tends to infinity as $n$ tends to infinity



Is there anything wrong with this proof?



thank you






real-analysis measure-theory proof-verification convergence sequence-of-function







share|cite|improve this question















This result seems too convenient and I feel like Folland would have used this to prove proposition 2.30 if this was true.



My "proof":



Let $f_n$ be Cauchy in measure.



By definition of $liminf f_n(x)$,
for every $n in mathbb{N}, x in X$ there exists an $m_{x,n}> n$ such that
$|liminf f_n(x) - f_m(x)| < epsilon / 2$



Now consider the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$}



By triangle inequality, $epsilon leq |f_n(x) -liminf f_n(x)| leq |f_n(x) - f_{m_{x,n}}(x)| + |f_{m_{x,n}}(x) - liminf f_n(x)|$



but since $|liminf f_n(x) - f_{m_{x,n}}(x)| leq epsilon / 2$, we must have that $ epsilon/2 leq |f_n(x) -liminf f_n(x)| $



therefore, the set



{$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} $subset$ {$xin X : |f_n(x) - f_{m_{x,n}}(x)| > epsilon/2$}



but the measure of the set on the right approaches to zero as $n$ tends to infinity (since $m_{x,n}> n$)



therefore the measure of the set {$xin X : |f_n(x) -liminf f_n(x)| > epsilon$} tends to infinity as $n$ tends to infinity



Is there anything wrong with this proof?



thank you






real-analysis measure-theory proof-verification convergence sequence-of-function




real-analysis measure-theory proof-verification convergence sequence-of-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 23:31

























asked Dec 2 '18 at 23:18









Ecotistician

31318




31318












  • It's liminf of what?
    – Bernard
    Dec 2 '18 at 23:24










  • sorry, lim inf fn(x). fixed.
    – Ecotistician
    Dec 2 '18 at 23:25












  • Use liminf (with a backslash) for the limit inferior.
    – Bernard
    Dec 2 '18 at 23:26






  • 1




    Yes, the proof is wrong. You only know that measure of ${x:|f_n(x)-f_m(x)| >epsilon} to 0$ as $n, m to infty$ and you cannot replace $m$ by an integer that depends on $x$.
    – Kavi Rama Murthy
    Dec 2 '18 at 23:37










  • If $liminf f_n(x)=infty$ or $-infty$ then you cannot achieve $epsilon$-closeness to $f_m(x)$ like you want to. Now, it actually happens to be true that $liminf f_n(x)in(-infty,infty)$ almost everywhere. But you would need to show that to be true. There may be other problems with your proof that I have not noticed yet.
    – Ben W
    Dec 2 '18 at 23:37


















  • It's liminf of what?
    – Bernard
    Dec 2 '18 at 23:24










  • sorry, lim inf fn(x). fixed.
    – Ecotistician
    Dec 2 '18 at 23:25












  • Use liminf (with a backslash) for the limit inferior.
    – Bernard
    Dec 2 '18 at 23:26






  • 1




    Yes, the proof is wrong. You only know that measure of ${x:|f_n(x)-f_m(x)| >epsilon} to 0$ as $n, m to infty$ and you cannot replace $m$ by an integer that depends on $x$.
    – Kavi Rama Murthy
    Dec 2 '18 at 23:37










  • If $liminf f_n(x)=infty$ or $-infty$ then you cannot achieve $epsilon$-closeness to $f_m(x)$ like you want to. Now, it actually happens to be true that $liminf f_n(x)in(-infty,infty)$ almost everywhere. But you would need to show that to be true. There may be other problems with your proof that I have not noticed yet.
    – Ben W
    Dec 2 '18 at 23:37
















It's liminf of what?
– Bernard
Dec 2 '18 at 23:24




It's liminf of what?
– Bernard
Dec 2 '18 at 23:24












sorry, lim inf fn(x). fixed.
– Ecotistician
Dec 2 '18 at 23:25






sorry, lim inf fn(x). fixed.
– Ecotistician
Dec 2 '18 at 23:25














Use liminf (with a backslash) for the limit inferior.
– Bernard
Dec 2 '18 at 23:26




Use liminf (with a backslash) for the limit inferior.
– Bernard
Dec 2 '18 at 23:26




1




1




Yes, the proof is wrong. You only know that measure of ${x:|f_n(x)-f_m(x)| >epsilon} to 0$ as $n, m to infty$ and you cannot replace $m$ by an integer that depends on $x$.
– Kavi Rama Murthy
Dec 2 '18 at 23:37




Yes, the proof is wrong. You only know that measure of ${x:|f_n(x)-f_m(x)| >epsilon} to 0$ as $n, m to infty$ and you cannot replace $m$ by an integer that depends on $x$.
– Kavi Rama Murthy
Dec 2 '18 at 23:37












If $liminf f_n(x)=infty$ or $-infty$ then you cannot achieve $epsilon$-closeness to $f_m(x)$ like you want to. Now, it actually happens to be true that $liminf f_n(x)in(-infty,infty)$ almost everywhere. But you would need to show that to be true. There may be other problems with your proof that I have not noticed yet.
– Ben W
Dec 2 '18 at 23:37




If $liminf f_n(x)=infty$ or $-infty$ then you cannot achieve $epsilon$-closeness to $f_m(x)$ like you want to. Now, it actually happens to be true that $liminf f_n(x)in(-infty,infty)$ almost everywhere. But you would need to show that to be true. There may be other problems with your proof that I have not noticed yet.
– Ben W
Dec 2 '18 at 23:37















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