Why don't we always consider complete measure spaces?
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Let $(Omega ,mathcal F,mathbb P)$ a probability space and let $X=(X_t)$ and $Y=(Y_t)$ two stochastic processes. I know for example that $X$ and $Y$ are indistinguishable if there is a set $N$ of measure $0$ s.t. for all $omega notin N$ we have $X_t=Y_t$ for all $t$, but we can't write $mathbb P{forall t, X_t=Y_t}$ since ${forall t, X_t=Y_t}$ may be not $mathcal F-$ measurable.
The thing is if $Y$ is a copy of $X$ and $(Omega ,mathcal F,mathbb P)$ is complete, then ${forall t,X_t=Y_t}$ is $mathcal F-$measurable.
I also know that each measure space can be completed by adding sets of measure.
Questions :
So, why don't we always work with complete measure space (since they can be always completed), and avoid for example the problem of the measurability of ${forall t, X_t=Y_t}$ if $Y$ is a copy of $X$ (or many other measurability problem) ?
In what working in a non complete measure space can be interesting, or at least more interesting than to work with it's completion ? (since a non complete measure space can always be completed).
Do you have an example where it's worth to work with the uncompleted measure space rather than with the completed space ?
probability
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
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add a comment |
$begingroup$
Let $(Omega ,mathcal F,mathbb P)$ a probability space and let $X=(X_t)$ and $Y=(Y_t)$ two stochastic processes. I know for example that $X$ and $Y$ are indistinguishable if there is a set $N$ of measure $0$ s.t. for all $omega notin N$ we have $X_t=Y_t$ for all $t$, but we can't write $mathbb P{forall t, X_t=Y_t}$ since ${forall t, X_t=Y_t}$ may be not $mathcal F-$ measurable.
The thing is if $Y$ is a copy of $X$ and $(Omega ,mathcal F,mathbb P)$ is complete, then ${forall t,X_t=Y_t}$ is $mathcal F-$measurable.
I also know that each measure space can be completed by adding sets of measure.
Questions :
So, why don't we always work with complete measure space (since they can be always completed), and avoid for example the problem of the measurability of ${forall t, X_t=Y_t}$ if $Y$ is a copy of $X$ (or many other measurability problem) ?
In what working in a non complete measure space can be interesting, or at least more interesting than to work with it's completion ? (since a non complete measure space can always be completed).
Do you have an example where it's worth to work with the uncompleted measure space rather than with the completed space ?
probability
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
Let $(Omega ,mathcal F,mathbb P)$ a probability space and let $X=(X_t)$ and $Y=(Y_t)$ two stochastic processes. I know for example that $X$ and $Y$ are indistinguishable if there is a set $N$ of measure $0$ s.t. for all $omega notin N$ we have $X_t=Y_t$ for all $t$, but we can't write $mathbb P{forall t, X_t=Y_t}$ since ${forall t, X_t=Y_t}$ may be not $mathcal F-$ measurable.
The thing is if $Y$ is a copy of $X$ and $(Omega ,mathcal F,mathbb P)$ is complete, then ${forall t,X_t=Y_t}$ is $mathcal F-$measurable.
I also know that each measure space can be completed by adding sets of measure.
Questions :
So, why don't we always work with complete measure space (since they can be always completed), and avoid for example the problem of the measurability of ${forall t, X_t=Y_t}$ if $Y$ is a copy of $X$ (or many other measurability problem) ?
In what working in a non complete measure space can be interesting, or at least more interesting than to work with it's completion ? (since a non complete measure space can always be completed).
Do you have an example where it's worth to work with the uncompleted measure space rather than with the completed space ?
probability
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
$endgroup$
Let $(Omega ,mathcal F,mathbb P)$ a probability space and let $X=(X_t)$ and $Y=(Y_t)$ two stochastic processes. I know for example that $X$ and $Y$ are indistinguishable if there is a set $N$ of measure $0$ s.t. for all $omega notin N$ we have $X_t=Y_t$ for all $t$, but we can't write $mathbb P{forall t, X_t=Y_t}$ since ${forall t, X_t=Y_t}$ may be not $mathcal F-$ measurable.
The thing is if $Y$ is a copy of $X$ and $(Omega ,mathcal F,mathbb P)$ is complete, then ${forall t,X_t=Y_t}$ is $mathcal F-$measurable.
I also know that each measure space can be completed by adding sets of measure.
Questions :
So, why don't we always work with complete measure space (since they can be always completed), and avoid for example the problem of the measurability of ${forall t, X_t=Y_t}$ if $Y$ is a copy of $X$ (or many other measurability problem) ?
In what working in a non complete measure space can be interesting, or at least more interesting than to work with it's completion ? (since a non complete measure space can always be completed).
Do you have an example where it's worth to work with the uncompleted measure space rather than with the completed space ?
probability
probability
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
asked Dec 13 '18 at 17:13
NewMathNewMath
4059
4059
add a comment |
add a comment |
1 Answer
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The main advantage I can think of is that the composition of two $(mathcal B,mathcal B)$-measurable functions are also $(mathcal B,mathcal B)$-measurable whereas the composition of two $(mathcal L,mathcal B)$-measurable functions need not be $(mathcal L,mathcal B)$-measurable. Of course, here $mathcal B$ is the family of Borel sets and $mathcal L$ is the family of Lebesgue measurable sets.
Many analysis books mean to say $(mathcal L,mathcal B)$-measurable when they say measurable. This is why those books state that we need a continuous function $varphi$ so that $varphicirc f$ is measurable, provided that $f$ is a measurable function.
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
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3
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Yes, good remark ! (+1) An other thing I would mention is that when $Omega $ is a separable topological space, the Borel $sigma -$algebra is countably generated.
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– Surb
Dec 13 '18 at 17:31
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BigbearZzz : Hey, very good remark ! It make sense to consider Borel $sigma -$algebra ! Now I really get it :) @Surb : I don't understand why it's a good thing, but I trust you ;)
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– NewMath
Dec 13 '18 at 17:56
1
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@NewMath: Don't worry, I'm sure you'll get it when you'll be a bit more used to it :-)
$endgroup$
– Surb
Dec 13 '18 at 18:00
add a comment |
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The main advantage I can think of is that the composition of two $(mathcal B,mathcal B)$-measurable functions are also $(mathcal B,mathcal B)$-measurable whereas the composition of two $(mathcal L,mathcal B)$-measurable functions need not be $(mathcal L,mathcal B)$-measurable. Of course, here $mathcal B$ is the family of Borel sets and $mathcal L$ is the family of Lebesgue measurable sets.
Many analysis books mean to say $(mathcal L,mathcal B)$-measurable when they say measurable. This is why those books state that we need a continuous function $varphi$ so that $varphicirc f$ is measurable, provided that $f$ is a measurable function.
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
$endgroup$
3
$begingroup$
Yes, good remark ! (+1) An other thing I would mention is that when $Omega $ is a separable topological space, the Borel $sigma -$algebra is countably generated.
$endgroup$
– Surb
Dec 13 '18 at 17:31
$begingroup$
BigbearZzz : Hey, very good remark ! It make sense to consider Borel $sigma -$algebra ! Now I really get it :) @Surb : I don't understand why it's a good thing, but I trust you ;)
$endgroup$
– NewMath
Dec 13 '18 at 17:56
1
$begingroup$
@NewMath: Don't worry, I'm sure you'll get it when you'll be a bit more used to it :-)
$endgroup$
– Surb
Dec 13 '18 at 18:00
add a comment |
$begingroup$
The main advantage I can think of is that the composition of two $(mathcal B,mathcal B)$-measurable functions are also $(mathcal B,mathcal B)$-measurable whereas the composition of two $(mathcal L,mathcal B)$-measurable functions need not be $(mathcal L,mathcal B)$-measurable. Of course, here $mathcal B$ is the family of Borel sets and $mathcal L$ is the family of Lebesgue measurable sets.
Many analysis books mean to say $(mathcal L,mathcal B)$-measurable when they say measurable. This is why those books state that we need a continuous function $varphi$ so that $varphicirc f$ is measurable, provided that $f$ is a measurable function.
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
$endgroup$
3
$begingroup$
Yes, good remark ! (+1) An other thing I would mention is that when $Omega $ is a separable topological space, the Borel $sigma -$algebra is countably generated.
$endgroup$
– Surb
Dec 13 '18 at 17:31
$begingroup$
BigbearZzz : Hey, very good remark ! It make sense to consider Borel $sigma -$algebra ! Now I really get it :) @Surb : I don't understand why it's a good thing, but I trust you ;)
$endgroup$
– NewMath
Dec 13 '18 at 17:56
1
$begingroup$
@NewMath: Don't worry, I'm sure you'll get it when you'll be a bit more used to it :-)
$endgroup$
– Surb
Dec 13 '18 at 18:00
add a comment |
$begingroup$
The main advantage I can think of is that the composition of two $(mathcal B,mathcal B)$-measurable functions are also $(mathcal B,mathcal B)$-measurable whereas the composition of two $(mathcal L,mathcal B)$-measurable functions need not be $(mathcal L,mathcal B)$-measurable. Of course, here $mathcal B$ is the family of Borel sets and $mathcal L$ is the family of Lebesgue measurable sets.
Many analysis books mean to say $(mathcal L,mathcal B)$-measurable when they say measurable. This is why those books state that we need a continuous function $varphi$ so that $varphicirc f$ is measurable, provided that $f$ is a measurable function.
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
$endgroup$
The main advantage I can think of is that the composition of two $(mathcal B,mathcal B)$-measurable functions are also $(mathcal B,mathcal B)$-measurable whereas the composition of two $(mathcal L,mathcal B)$-measurable functions need not be $(mathcal L,mathcal B)$-measurable. Of course, here $mathcal B$ is the family of Borel sets and $mathcal L$ is the family of Lebesgue measurable sets.
Many analysis books mean to say $(mathcal L,mathcal B)$-measurable when they say measurable. This is why those books state that we need a continuous function $varphi$ so that $varphicirc f$ is measurable, provided that $f$ is a measurable function.
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
edited Dec 13 '18 at 18:30
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
answered Dec 13 '18 at 17:26
BigbearZzzBigbearZzz
8,88821652
8,88821652
3
$begingroup$
Yes, good remark ! (+1) An other thing I would mention is that when $Omega $ is a separable topological space, the Borel $sigma -$algebra is countably generated.
$endgroup$
– Surb
Dec 13 '18 at 17:31
$begingroup$
BigbearZzz : Hey, very good remark ! It make sense to consider Borel $sigma -$algebra ! Now I really get it :) @Surb : I don't understand why it's a good thing, but I trust you ;)
$endgroup$
– NewMath
Dec 13 '18 at 17:56
1
$begingroup$
@NewMath: Don't worry, I'm sure you'll get it when you'll be a bit more used to it :-)
$endgroup$
– Surb
Dec 13 '18 at 18:00
add a comment |
3
$begingroup$
Yes, good remark ! (+1) An other thing I would mention is that when $Omega $ is a separable topological space, the Borel $sigma -$algebra is countably generated.
$endgroup$
– Surb
Dec 13 '18 at 17:31
$begingroup$
BigbearZzz : Hey, very good remark ! It make sense to consider Borel $sigma -$algebra ! Now I really get it :) @Surb : I don't understand why it's a good thing, but I trust you ;)
$endgroup$
– NewMath
Dec 13 '18 at 17:56
1
$begingroup$
@NewMath: Don't worry, I'm sure you'll get it when you'll be a bit more used to it :-)
$endgroup$
– Surb
Dec 13 '18 at 18:00
3
3
$begingroup$
Yes, good remark ! (+1) An other thing I would mention is that when $Omega $ is a separable topological space, the Borel $sigma -$algebra is countably generated.
$endgroup$
– Surb
Dec 13 '18 at 17:31
$begingroup$
Yes, good remark ! (+1) An other thing I would mention is that when $Omega $ is a separable topological space, the Borel $sigma -$algebra is countably generated.
$endgroup$
– Surb
Dec 13 '18 at 17:31
$begingroup$
BigbearZzz : Hey, very good remark ! It make sense to consider Borel $sigma -$algebra ! Now I really get it :) @Surb : I don't understand why it's a good thing, but I trust you ;)
$endgroup$
– NewMath
Dec 13 '18 at 17:56
$begingroup$
BigbearZzz : Hey, very good remark ! It make sense to consider Borel $sigma -$algebra ! Now I really get it :) @Surb : I don't understand why it's a good thing, but I trust you ;)
$endgroup$
– NewMath
Dec 13 '18 at 17:56
1
1
$begingroup$
@NewMath: Don't worry, I'm sure you'll get it when you'll be a bit more used to it :-)
$endgroup$
– Surb
Dec 13 '18 at 18:00
$begingroup$
@NewMath: Don't worry, I'm sure you'll get it when you'll be a bit more used to it :-)
$endgroup$
– Surb
Dec 13 '18 at 18:00
add a comment |
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