Mixing
Converge in measure implies converge a.e if $f_n$ are monotone
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Let ${f_n}$ be a sequence of monotone functions such that $f_n$ converges in measure to $f$. Is it true that $f_n$ converges to $f$ a.e.?
I am sure it has a sub-sequence that converges to $f$ a.e. and intuitively it seems that it must be true but i am not able to write mathematically, could you please give me some hint.
real-analysis measure-theory
edited Jan 30 at 17:52
Mars Plastic
1,477122
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
$endgroup$
add a comment |
$begingroup$
Let ${f_n}$ be a sequence of monotone functions such that $f_n$ converges in measure to $f$. Is it true that $f_n$ converges to $f$ a.e.?
I am sure it has a sub-sequence that converges to $f$ a.e. and intuitively it seems that it must be true but i am not able to write mathematically, could you please give me some hint.
real-analysis measure-theory
edited Jan 30 at 17:52
Mars Plastic
1,477122
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
$endgroup$
2
$begingroup$
Is $f_n$ monotone wrt $x$ or wrt $n$? If wrt $n$ then this is trivial, $f_n$ converges everywhere (possibly to $pm infty$) without assuming it converges in measure. If wrt $x$ then there's a bit more to do. Also, is the measure space in question finite?
$endgroup$
– Ian
Jul 16 '14 at 15:45
$begingroup$
yes, you are correct, it must be monotone in $x$. About space it doesn't specify whether finite or not.
$endgroup$
– Toeplitz
Jul 16 '14 at 15:52
$begingroup$
Could you do it if the $f_n$ were continuous? The monotonicity implies the $A = { x : (exists n) f_n text{ is not continuous at } x }$ is countable, hence measure zero. That might help.
$endgroup$
– Ian
Jul 16 '14 at 15:57
add a comment |
$begingroup$
Let ${f_n}$ be a sequence of monotone functions such that $f_n$ converges in measure to $f$. Is it true that $f_n$ converges to $f$ a.e.?
I am sure it has a sub-sequence that converges to $f$ a.e. and intuitively it seems that it must be true but i am not able to write mathematically, could you please give me some hint.
real-analysis measure-theory
edited Jan 30 at 17:52
Mars Plastic
1,477122
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
$endgroup$
Let ${f_n}$ be a sequence of monotone functions such that $f_n$ converges in measure to $f$. Is it true that $f_n$ converges to $f$ a.e.?
I am sure it has a sub-sequence that converges to $f$ a.e. and intuitively it seems that it must be true but i am not able to write mathematically, could you please give me some hint.
real-analysis measure-theory
real-analysis measure-theory
edited Jan 30 at 17:52
Mars Plastic
1,477122
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
edited Jan 30 at 17:52
Mars Plastic
1,477122
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
edited Jan 30 at 17:52
Mars Plastic
1,477122
edited Jan 30 at 17:52
Mars Plastic
1,477122
edited Jan 30 at 17:52
Mars Plastic
1,477122
1,477122
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
asked Jul 16 '14 at 15:31
ToeplitzToeplitz
673513
673513
2
$begingroup$
Is $f_n$ monotone wrt $x$ or wrt $n$? If wrt $n$ then this is trivial, $f_n$ converges everywhere (possibly to $pm infty$) without assuming it converges in measure. If wrt $x$ then there's a bit more to do. Also, is the measure space in question finite?
$endgroup$
– Ian
Jul 16 '14 at 15:45
$begingroup$
yes, you are correct, it must be monotone in $x$. About space it doesn't specify whether finite or not.
$endgroup$
– Toeplitz
Jul 16 '14 at 15:52
$begingroup$
Could you do it if the $f_n$ were continuous? The monotonicity implies the $A = { x : (exists n) f_n text{ is not continuous at } x }$ is countable, hence measure zero. That might help.
$endgroup$
– Ian
Jul 16 '14 at 15:57
add a comment |
2
$begingroup$
Is $f_n$ monotone wrt $x$ or wrt $n$? If wrt $n$ then this is trivial, $f_n$ converges everywhere (possibly to $pm infty$) without assuming it converges in measure. If wrt $x$ then there's a bit more to do. Also, is the measure space in question finite?
$endgroup$
– Ian
Jul 16 '14 at 15:45
$begingroup$
yes, you are correct, it must be monotone in $x$. About space it doesn't specify whether finite or not.
$endgroup$
– Toeplitz
Jul 16 '14 at 15:52
$begingroup$
Could you do it if the $f_n$ were continuous? The monotonicity implies the $A = { x : (exists n) f_n text{ is not continuous at } x }$ is countable, hence measure zero. That might help.
$endgroup$
– Ian
Jul 16 '14 at 15:57
2
2
$begingroup$
Is $f_n$ monotone wrt $x$ or wrt $n$? If wrt $n$ then this is trivial, $f_n$ converges everywhere (possibly to $pm infty$) without assuming it converges in measure. If wrt $x$ then there's a bit more to do. Also, is the measure space in question finite?
$endgroup$
– Ian
Jul 16 '14 at 15:45
$begingroup$
Is $f_n$ monotone wrt $x$ or wrt $n$? If wrt $n$ then this is trivial, $f_n$ converges everywhere (possibly to $pm infty$) without assuming it converges in measure. If wrt $x$ then there's a bit more to do. Also, is the measure space in question finite?
$endgroup$
– Ian
Jul 16 '14 at 15:45
$begingroup$
yes, you are correct, it must be monotone in $x$. About space it doesn't specify whether finite or not.
$endgroup$
– Toeplitz
Jul 16 '14 at 15:52
$begingroup$
yes, you are correct, it must be monotone in $x$. About space it doesn't specify whether finite or not.
$endgroup$
– Toeplitz
Jul 16 '14 at 15:52
$begingroup$
Could you do it if the $f_n$ were continuous? The monotonicity implies the $A = { x : (exists n) f_n text{ is not continuous at } x }$ is countable, hence measure zero. That might help.
$endgroup$
– Ian
Jul 16 '14 at 15:57
$begingroup$
Could you do it if the $f_n$ were continuous? The monotonicity implies the $A = { x : (exists n) f_n text{ is not continuous at } x }$ is countable, hence measure zero. That might help.
$endgroup$
– Ian
Jul 16 '14 at 15:57
add a comment |
1 Answer
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$begingroup$
If $(f_n)$ converges to $f$ in measure, there is a subsequence converging to f almost everywhere. So there exist subsequence $(n_i )subsetmathbb{N}$ and a subsetset $Asubset X$ such that $mu (A)=0 $ and $$forall_{tin Xsetminus A } lim_{itoinfty } (f(t)-f_{n_i } (t) )=0$$ but $$f(t)-f_{n_i } (t)geq f(t) -f_k (t) mbox{ for } kgeq n_i $$ hence the sequence $(f_n )$ converges almost everywhere.
$endgroup$
2
$begingroup$
As was clarified in the comments, the monotonicity is with respect to $x$, not $n$; we have a "sequence of monotone functions", not a "monotone sequence of functions". When the monotonicity is with respect to $n$ there is no measure theory required at all.
$endgroup$
– Ian
Jul 16 '14 at 15:58
add a comment |
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1 Answer
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$begingroup$
If $(f_n)$ converges to $f$ in measure, there is a subsequence converging to f almost everywhere. So there exist subsequence $(n_i )subsetmathbb{N}$ and a subsetset $Asubset X$ such that $mu (A)=0 $ and $$forall_{tin Xsetminus A } lim_{itoinfty } (f(t)-f_{n_i } (t) )=0$$ but $$f(t)-f_{n_i } (t)geq f(t) -f_k (t) mbox{ for } kgeq n_i $$ hence the sequence $(f_n )$ converges almost everywhere.
$endgroup$
2
$begingroup$
As was clarified in the comments, the monotonicity is with respect to $x$, not $n$; we have a "sequence of monotone functions", not a "monotone sequence of functions". When the monotonicity is with respect to $n$ there is no measure theory required at all.
$endgroup$
– Ian
Jul 16 '14 at 15:58
add a comment |
$begingroup$
If $(f_n)$ converges to $f$ in measure, there is a subsequence converging to f almost everywhere. So there exist subsequence $(n_i )subsetmathbb{N}$ and a subsetset $Asubset X$ such that $mu (A)=0 $ and $$forall_{tin Xsetminus A } lim_{itoinfty } (f(t)-f_{n_i } (t) )=0$$ but $$f(t)-f_{n_i } (t)geq f(t) -f_k (t) mbox{ for } kgeq n_i $$ hence the sequence $(f_n )$ converges almost everywhere.
$endgroup$
2
$begingroup$
As was clarified in the comments, the monotonicity is with respect to $x$, not $n$; we have a "sequence of monotone functions", not a "monotone sequence of functions". When the monotonicity is with respect to $n$ there is no measure theory required at all.
$endgroup$
– Ian
Jul 16 '14 at 15:58
add a comment |
$begingroup$
If $(f_n)$ converges to $f$ in measure, there is a subsequence converging to f almost everywhere. So there exist subsequence $(n_i )subsetmathbb{N}$ and a subsetset $Asubset X$ such that $mu (A)=0 $ and $$forall_{tin Xsetminus A } lim_{itoinfty } (f(t)-f_{n_i } (t) )=0$$ but $$f(t)-f_{n_i } (t)geq f(t) -f_k (t) mbox{ for } kgeq n_i $$ hence the sequence $(f_n )$ converges almost everywhere.
$endgroup$
If $(f_n)$ converges to $f$ in measure, there is a subsequence converging to f almost everywhere. So there exist subsequence $(n_i )subsetmathbb{N}$ and a subsetset $Asubset X$ such that $mu (A)=0 $ and $$forall_{tin Xsetminus A } lim_{itoinfty } (f(t)-f_{n_i } (t) )=0$$ but $$f(t)-f_{n_i } (t)geq f(t) -f_k (t) mbox{ for } kgeq n_i $$ hence the sequence $(f_n )$ converges almost everywhere.
answered Jul 16 '14 at 15:57
user110661
2
$begingroup$
As was clarified in the comments, the monotonicity is with respect to $x$, not $n$; we have a "sequence of monotone functions", not a "monotone sequence of functions". When the monotonicity is with respect to $n$ there is no measure theory required at all.
$endgroup$
– Ian
Jul 16 '14 at 15:58
add a comment |
2
$begingroup$
As was clarified in the comments, the monotonicity is with respect to $x$, not $n$; we have a "sequence of monotone functions", not a "monotone sequence of functions". When the monotonicity is with respect to $n$ there is no measure theory required at all.
$endgroup$
– Ian
Jul 16 '14 at 15:58
2
2
$begingroup$
As was clarified in the comments, the monotonicity is with respect to $x$, not $n$; we have a "sequence of monotone functions", not a "monotone sequence of functions". When the monotonicity is with respect to $n$ there is no measure theory required at all.
$endgroup$
– Ian
Jul 16 '14 at 15:58
$begingroup$
As was clarified in the comments, the monotonicity is with respect to $x$, not $n$; we have a "sequence of monotone functions", not a "monotone sequence of functions". When the monotonicity is with respect to $n$ there is no measure theory required at all.
$endgroup$
– Ian
Jul 16 '14 at 15:58
add a comment |
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$begingroup$
Is $f_n$ monotone wrt $x$ or wrt $n$? If wrt $n$ then this is trivial, $f_n$ converges everywhere (possibly to $pm infty$) without assuming it converges in measure. If wrt $x$ then there's a bit more to do. Also, is the measure space in question finite?
$endgroup$
– Ian
Jul 16 '14 at 15:45
$begingroup$
yes, you are correct, it must be monotone in $x$. About space it doesn't specify whether finite or not.
$endgroup$
– Toeplitz
Jul 16 '14 at 15:52
$begingroup$
Could you do it if the $f_n$ were continuous? The monotonicity implies the $A = { x : (exists n) f_n text{ is not continuous at } x }$ is countable, hence measure zero. That might help.
$endgroup$
– Ian
Jul 16 '14 at 15:57