For which measures does the monotone convergence theorem hold?
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The standard proof of the monotone convergence theorem for the Lebesgue measure picks a simple function $phi$ that is uniformly bounded above by the pointwise limit $f$ of the monotone increasing sequence of positive (Lebesgue) measurable functions ${f_n }$ in question; fixing a $k in (0,1)$, sets of the form $E_n = {x in mathbb{R} : f_n(x) geq kphi(x)}$ are claimed to be an increasing sequence of measurable sets that cover the entire space (the reals). (Sneaking in a quick question here: why do we use $k$ at all? Is it so that we can always 'squeeze in' elements in $E_n$ in case $phi=f$?)
I think the claim that the sets are measurable relies on completeness of the Lebesgue measure; could someone confirm this? Can this proof be adapted to non-complete measures (say, the Borel measure)? For what kind of measures on the reals does the convergence theorem hold? Is there an analogue of the theorem for measures on other ordered fields possessing the least upper bound property?
measure-theory
asked Nov 15 at 11:34
SystematicDisintegration
274211
add a comment |
up vote
1
down vote
favorite
The standard proof of the monotone convergence theorem for the Lebesgue measure picks a simple function $phi$ that is uniformly bounded above by the pointwise limit $f$ of the monotone increasing sequence of positive (Lebesgue) measurable functions ${f_n }$ in question; fixing a $k in (0,1)$, sets of the form $E_n = {x in mathbb{R} : f_n(x) geq kphi(x)}$ are claimed to be an increasing sequence of measurable sets that cover the entire space (the reals). (Sneaking in a quick question here: why do we use $k$ at all? Is it so that we can always 'squeeze in' elements in $E_n$ in case $phi=f$?)
I think the claim that the sets are measurable relies on completeness of the Lebesgue measure; could someone confirm this? Can this proof be adapted to non-complete measures (say, the Borel measure)? For what kind of measures on the reals does the convergence theorem hold? Is there an analogue of the theorem for measures on other ordered fields possessing the least upper bound property?
measure-theory
asked Nov 15 at 11:34
SystematicDisintegration
274211
Actually I do not recognize your description of the proof. Can you give a link to it? Have you taken a look here already?
– drhab
Nov 15 at 12:08
You will find the exact description in the link in the proof section of the case with the Lebesgue integral -> 'proof of the theorem' -> independent proof (step 3).
– SystematicDisintegration
Nov 15 at 12:14
Thank you. But nowhere I can find that it should only concern the Lebesgue measure (or a complete measure).
– drhab
Nov 15 at 12:19
Right; I meant to seek confirmation on the same.
– SystematicDisintegration
Nov 15 at 12:23
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The standard proof of the monotone convergence theorem for the Lebesgue measure picks a simple function $phi$ that is uniformly bounded above by the pointwise limit $f$ of the monotone increasing sequence of positive (Lebesgue) measurable functions ${f_n }$ in question; fixing a $k in (0,1)$, sets of the form $E_n = {x in mathbb{R} : f_n(x) geq kphi(x)}$ are claimed to be an increasing sequence of measurable sets that cover the entire space (the reals). (Sneaking in a quick question here: why do we use $k$ at all? Is it so that we can always 'squeeze in' elements in $E_n$ in case $phi=f$?)
I think the claim that the sets are measurable relies on completeness of the Lebesgue measure; could someone confirm this? Can this proof be adapted to non-complete measures (say, the Borel measure)? For what kind of measures on the reals does the convergence theorem hold? Is there an analogue of the theorem for measures on other ordered fields possessing the least upper bound property?
measure-theory
asked Nov 15 at 11:34
SystematicDisintegration
274211
The standard proof of the monotone convergence theorem for the Lebesgue measure picks a simple function $phi$ that is uniformly bounded above by the pointwise limit $f$ of the monotone increasing sequence of positive (Lebesgue) measurable functions ${f_n }$ in question; fixing a $k in (0,1)$, sets of the form $E_n = {x in mathbb{R} : f_n(x) geq kphi(x)}$ are claimed to be an increasing sequence of measurable sets that cover the entire space (the reals). (Sneaking in a quick question here: why do we use $k$ at all? Is it so that we can always 'squeeze in' elements in $E_n$ in case $phi=f$?)
I think the claim that the sets are measurable relies on completeness of the Lebesgue measure; could someone confirm this? Can this proof be adapted to non-complete measures (say, the Borel measure)? For what kind of measures on the reals does the convergence theorem hold? Is there an analogue of the theorem for measures on other ordered fields possessing the least upper bound property?
measure-theory
measure-theory
asked Nov 15 at 11:34
SystematicDisintegration
274211
asked Nov 15 at 11:34
SystematicDisintegration
274211
edited Nov 15 at 12:00
asked Nov 15 at 11:34
SystematicDisintegration
274211
asked Nov 15 at 11:34
SystematicDisintegration
274211
asked Nov 15 at 11:34
SystematicDisintegration
274211
274211
Actually I do not recognize your description of the proof. Can you give a link to it? Have you taken a look here already?
– drhab
Nov 15 at 12:08
You will find the exact description in the link in the proof section of the case with the Lebesgue integral -> 'proof of the theorem' -> independent proof (step 3).
– SystematicDisintegration
Nov 15 at 12:14
Thank you. But nowhere I can find that it should only concern the Lebesgue measure (or a complete measure).
– drhab
Nov 15 at 12:19
Right; I meant to seek confirmation on the same.
– SystematicDisintegration
Nov 15 at 12:23
add a comment |
Actually I do not recognize your description of the proof. Can you give a link to it? Have you taken a look here already?
– drhab
Nov 15 at 12:08
You will find the exact description in the link in the proof section of the case with the Lebesgue integral -> 'proof of the theorem' -> independent proof (step 3).
– SystematicDisintegration
Nov 15 at 12:14
Thank you. But nowhere I can find that it should only concern the Lebesgue measure (or a complete measure).
– drhab
Nov 15 at 12:19
Right; I meant to seek confirmation on the same.
– SystematicDisintegration
Nov 15 at 12:23
Actually I do not recognize your description of the proof. Can you give a link to it? Have you taken a look here already?
– drhab
Nov 15 at 12:08
Actually I do not recognize your description of the proof. Can you give a link to it? Have you taken a look here already?
– drhab
Nov 15 at 12:08
You will find the exact description in the link in the proof section of the case with the Lebesgue integral -> 'proof of the theorem' -> independent proof (step 3).
– SystematicDisintegration
Nov 15 at 12:14
You will find the exact description in the link in the proof section of the case with the Lebesgue integral -> 'proof of the theorem' -> independent proof (step 3).
– SystematicDisintegration
Nov 15 at 12:14
Thank you. But nowhere I can find that it should only concern the Lebesgue measure (or a complete measure).
– drhab
Nov 15 at 12:19
Thank you. But nowhere I can find that it should only concern the Lebesgue measure (or a complete measure).
– drhab
Nov 15 at 12:19
Right; I meant to seek confirmation on the same.
– SystematicDisintegration
Nov 15 at 12:23
Right; I meant to seek confirmation on the same.
– SystematicDisintegration
Nov 15 at 12:23
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The monotone convergence theorem holds for any measure space: wiki
The real numbers are (up to order isomorphism) the only ordered field with the least upper bound property (this property is called "Dedekind completeness") This is a well-known fact (wiki) and there is a proof in an appendix of Spivak's "Calculus".
Sneak answer: if you don't use in $k$ then the sets do not necessarily cover $mathbb{R}$: let $g$ be the characteristic function of $[0,1]$. What if $f_n=(1-1/n)g$ and $phi=g$? Also, the proof should not require completeness of the Lebesgue $sigma$-algebra at any point.
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
If we disallow $phi$ from being the limit of the $f_n$, i.e. $f$ (which is equivalent to demanding the existence of a simple function strictly less than $f$ so that for a sufficiently large $m$, $f_m geq phi$), then can we discard the use of $k$?
– SystematicDisintegration
Nov 15 at 12:27
@SystematicDisintegration I assume you mean $phi(x)<f(x)$ whenever $f(x)neq 0$. In this case, yes, we don't need to use $k$. But there is a trade-off, since we would have to prove that the integral of $f$ is the supremum of the integrals of such $phi$. This argument should actually be somewhat hidden in the proof of the monotone convergence, if it is the one I have in mind.
– Luiz Cordeiro
Nov 16 at 12:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The monotone convergence theorem holds for any measure space: wiki
The real numbers are (up to order isomorphism) the only ordered field with the least upper bound property (this property is called "Dedekind completeness") This is a well-known fact (wiki) and there is a proof in an appendix of Spivak's "Calculus".
Sneak answer: if you don't use in $k$ then the sets do not necessarily cover $mathbb{R}$: let $g$ be the characteristic function of $[0,1]$. What if $f_n=(1-1/n)g$ and $phi=g$? Also, the proof should not require completeness of the Lebesgue $sigma$-algebra at any point.
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
If we disallow $phi$ from being the limit of the $f_n$, i.e. $f$ (which is equivalent to demanding the existence of a simple function strictly less than $f$ so that for a sufficiently large $m$, $f_m geq phi$), then can we discard the use of $k$?
– SystematicDisintegration
Nov 15 at 12:27
@SystematicDisintegration I assume you mean $phi(x)<f(x)$ whenever $f(x)neq 0$. In this case, yes, we don't need to use $k$. But there is a trade-off, since we would have to prove that the integral of $f$ is the supremum of the integrals of such $phi$. This argument should actually be somewhat hidden in the proof of the monotone convergence, if it is the one I have in mind.
– Luiz Cordeiro
Nov 16 at 12:19
add a comment |
up vote
2
down vote
accepted
The monotone convergence theorem holds for any measure space: wiki
The real numbers are (up to order isomorphism) the only ordered field with the least upper bound property (this property is called "Dedekind completeness") This is a well-known fact (wiki) and there is a proof in an appendix of Spivak's "Calculus".
Sneak answer: if you don't use in $k$ then the sets do not necessarily cover $mathbb{R}$: let $g$ be the characteristic function of $[0,1]$. What if $f_n=(1-1/n)g$ and $phi=g$? Also, the proof should not require completeness of the Lebesgue $sigma$-algebra at any point.
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
If we disallow $phi$ from being the limit of the $f_n$, i.e. $f$ (which is equivalent to demanding the existence of a simple function strictly less than $f$ so that for a sufficiently large $m$, $f_m geq phi$), then can we discard the use of $k$?
– SystematicDisintegration
Nov 15 at 12:27
@SystematicDisintegration I assume you mean $phi(x)<f(x)$ whenever $f(x)neq 0$. In this case, yes, we don't need to use $k$. But there is a trade-off, since we would have to prove that the integral of $f$ is the supremum of the integrals of such $phi$. This argument should actually be somewhat hidden in the proof of the monotone convergence, if it is the one I have in mind.
– Luiz Cordeiro
Nov 16 at 12:19
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The monotone convergence theorem holds for any measure space: wiki
The real numbers are (up to order isomorphism) the only ordered field with the least upper bound property (this property is called "Dedekind completeness") This is a well-known fact (wiki) and there is a proof in an appendix of Spivak's "Calculus".
Sneak answer: if you don't use in $k$ then the sets do not necessarily cover $mathbb{R}$: let $g$ be the characteristic function of $[0,1]$. What if $f_n=(1-1/n)g$ and $phi=g$? Also, the proof should not require completeness of the Lebesgue $sigma$-algebra at any point.
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
The monotone convergence theorem holds for any measure space: wiki
The real numbers are (up to order isomorphism) the only ordered field with the least upper bound property (this property is called "Dedekind completeness") This is a well-known fact (wiki) and there is a proof in an appendix of Spivak's "Calculus".
Sneak answer: if you don't use in $k$ then the sets do not necessarily cover $mathbb{R}$: let $g$ be the characteristic function of $[0,1]$. What if $f_n=(1-1/n)g$ and $phi=g$? Also, the proof should not require completeness of the Lebesgue $sigma$-algebra at any point.
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
edited Nov 15 at 12:23
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
answered Nov 15 at 12:16
Luiz Cordeiro
12.4k1143
12.4k1143
If we disallow $phi$ from being the limit of the $f_n$, i.e. $f$ (which is equivalent to demanding the existence of a simple function strictly less than $f$ so that for a sufficiently large $m$, $f_m geq phi$), then can we discard the use of $k$?
– SystematicDisintegration
Nov 15 at 12:27
@SystematicDisintegration I assume you mean $phi(x)<f(x)$ whenever $f(x)neq 0$. In this case, yes, we don't need to use $k$. But there is a trade-off, since we would have to prove that the integral of $f$ is the supremum of the integrals of such $phi$. This argument should actually be somewhat hidden in the proof of the monotone convergence, if it is the one I have in mind.
– Luiz Cordeiro
Nov 16 at 12:19
add a comment |
If we disallow $phi$ from being the limit of the $f_n$, i.e. $f$ (which is equivalent to demanding the existence of a simple function strictly less than $f$ so that for a sufficiently large $m$, $f_m geq phi$), then can we discard the use of $k$?
– SystematicDisintegration
Nov 15 at 12:27
@SystematicDisintegration I assume you mean $phi(x)<f(x)$ whenever $f(x)neq 0$. In this case, yes, we don't need to use $k$. But there is a trade-off, since we would have to prove that the integral of $f$ is the supremum of the integrals of such $phi$. This argument should actually be somewhat hidden in the proof of the monotone convergence, if it is the one I have in mind.
– Luiz Cordeiro
Nov 16 at 12:19
If we disallow $phi$ from being the limit of the $f_n$, i.e. $f$ (which is equivalent to demanding the existence of a simple function strictly less than $f$ so that for a sufficiently large $m$, $f_m geq phi$), then can we discard the use of $k$?
– SystematicDisintegration
Nov 15 at 12:27
If we disallow $phi$ from being the limit of the $f_n$, i.e. $f$ (which is equivalent to demanding the existence of a simple function strictly less than $f$ so that for a sufficiently large $m$, $f_m geq phi$), then can we discard the use of $k$?
– SystematicDisintegration
Nov 15 at 12:27
@SystematicDisintegration I assume you mean $phi(x)<f(x)$ whenever $f(x)neq 0$. In this case, yes, we don't need to use $k$. But there is a trade-off, since we would have to prove that the integral of $f$ is the supremum of the integrals of such $phi$. This argument should actually be somewhat hidden in the proof of the monotone convergence, if it is the one I have in mind.
– Luiz Cordeiro
Nov 16 at 12:19
@SystematicDisintegration I assume you mean $phi(x)<f(x)$ whenever $f(x)neq 0$. In this case, yes, we don't need to use $k$. But there is a trade-off, since we would have to prove that the integral of $f$ is the supremum of the integrals of such $phi$. This argument should actually be somewhat hidden in the proof of the monotone convergence, if it is the one I have in mind.
– Luiz Cordeiro
Nov 16 at 12:19
add a comment |
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Actually I do not recognize your description of the proof. Can you give a link to it? Have you taken a look here already?
– drhab
Nov 15 at 12:08
You will find the exact description in the link in the proof section of the case with the Lebesgue integral -> 'proof of the theorem' -> independent proof (step 3).
– SystematicDisintegration
Nov 15 at 12:14
Thank you. But nowhere I can find that it should only concern the Lebesgue measure (or a complete measure).
– drhab
Nov 15 at 12:19
Right; I meant to seek confirmation on the same.
– SystematicDisintegration
Nov 15 at 12:23