Mixing
Help on the two “separate” definitions of Product Measure
$begingroup$
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be $sigma-$finite measures, then there exists one $sigma-$finite measure $pi:=mu_{1}otimes mu_{2}$ on $mathcal{A}_{1} otimes mathcal{A}_{2}$ (i.e. the product measure) such that
$1.$ $mu_{1}otimes mu_{2}(A_{1}times A_{2})=mu_{1}(A_{1})mu_{2}(A_{2})$ for any $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$
$2.$ Then $forall A in mathcal{A}_{1} otimes mathcal{A}_{2}:int_{X_{1}}mu_{2}(A_{x_{1}})dmu_{1}(x_{1})=int_{X_{2}}mu_{1}(A_{x_{2}})dmu_{2}(x_{2})$
I struggle to find how $1.$ and $2.$ are related to each other. I mean the only difference I see is that in $1.$ we are looking at $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$ rather than $A in mathcal{A}_{1} otimes mathcal{A}_{2}$. What important concept am I missing here? Is $1.$ simply a special case of $2.$ since $mathcal{A}_{1}times mathcal{A}_{2}subseteq mathcal{A}_{1} otimes mathcal{A}_{2}$, as $mathcal{A}_{1} otimes mathcal{A}_{2}$ is the smallest $sigma-$Algebra such that the projection map is measurable?
real-analysis measure-theory lebesgue-measure
asked Jan 18 at 15:06
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be $sigma-$finite measures, then there exists one $sigma-$finite measure $pi:=mu_{1}otimes mu_{2}$ on $mathcal{A}_{1} otimes mathcal{A}_{2}$ (i.e. the product measure) such that
$1.$ $mu_{1}otimes mu_{2}(A_{1}times A_{2})=mu_{1}(A_{1})mu_{2}(A_{2})$ for any $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$
$2.$ Then $forall A in mathcal{A}_{1} otimes mathcal{A}_{2}:int_{X_{1}}mu_{2}(A_{x_{1}})dmu_{1}(x_{1})=int_{X_{2}}mu_{1}(A_{x_{2}})dmu_{2}(x_{2})$
I struggle to find how $1.$ and $2.$ are related to each other. I mean the only difference I see is that in $1.$ we are looking at $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$ rather than $A in mathcal{A}_{1} otimes mathcal{A}_{2}$. What important concept am I missing here? Is $1.$ simply a special case of $2.$ since $mathcal{A}_{1}times mathcal{A}_{2}subseteq mathcal{A}_{1} otimes mathcal{A}_{2}$, as $mathcal{A}_{1} otimes mathcal{A}_{2}$ is the smallest $sigma-$Algebra such that the projection map is measurable?
real-analysis measure-theory lebesgue-measure
asked Jan 18 at 15:06
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be $sigma-$finite measures, then there exists one $sigma-$finite measure $pi:=mu_{1}otimes mu_{2}$ on $mathcal{A}_{1} otimes mathcal{A}_{2}$ (i.e. the product measure) such that
$1.$ $mu_{1}otimes mu_{2}(A_{1}times A_{2})=mu_{1}(A_{1})mu_{2}(A_{2})$ for any $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$
$2.$ Then $forall A in mathcal{A}_{1} otimes mathcal{A}_{2}:int_{X_{1}}mu_{2}(A_{x_{1}})dmu_{1}(x_{1})=int_{X_{2}}mu_{1}(A_{x_{2}})dmu_{2}(x_{2})$
I struggle to find how $1.$ and $2.$ are related to each other. I mean the only difference I see is that in $1.$ we are looking at $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$ rather than $A in mathcal{A}_{1} otimes mathcal{A}_{2}$. What important concept am I missing here? Is $1.$ simply a special case of $2.$ since $mathcal{A}_{1}times mathcal{A}_{2}subseteq mathcal{A}_{1} otimes mathcal{A}_{2}$, as $mathcal{A}_{1} otimes mathcal{A}_{2}$ is the smallest $sigma-$Algebra such that the projection map is measurable?
real-analysis measure-theory lebesgue-measure
asked Jan 18 at 15:06
SABOYSABOY
656311
$endgroup$
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be $sigma-$finite measures, then there exists one $sigma-$finite measure $pi:=mu_{1}otimes mu_{2}$ on $mathcal{A}_{1} otimes mathcal{A}_{2}$ (i.e. the product measure) such that
$1.$ $mu_{1}otimes mu_{2}(A_{1}times A_{2})=mu_{1}(A_{1})mu_{2}(A_{2})$ for any $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$
$2.$ Then $forall A in mathcal{A}_{1} otimes mathcal{A}_{2}:int_{X_{1}}mu_{2}(A_{x_{1}})dmu_{1}(x_{1})=int_{X_{2}}mu_{1}(A_{x_{2}})dmu_{2}(x_{2})$
I struggle to find how $1.$ and $2.$ are related to each other. I mean the only difference I see is that in $1.$ we are looking at $A_{1} in mathcal{A}_{1}, A_{2} in mathcal{A}_{2}$ rather than $A in mathcal{A}_{1} otimes mathcal{A}_{2}$. What important concept am I missing here? Is $1.$ simply a special case of $2.$ since $mathcal{A}_{1}times mathcal{A}_{2}subseteq mathcal{A}_{1} otimes mathcal{A}_{2}$, as $mathcal{A}_{1} otimes mathcal{A}_{2}$ is the smallest $sigma-$Algebra such that the projection map is measurable?
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Jan 18 at 15:06
SABOYSABOY
656311
asked Jan 18 at 15:06
SABOYSABOY
656311
asked Jan 18 at 15:06
SABOYSABOY
656311
asked Jan 18 at 15:06
SABOYSABOY
656311
asked Jan 18 at 15:06
SABOYSABOY
656311
656311
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The actual product measure is given in (2).
If a measure on $mathcal{A_1} otimes mathcal{A_2}$ satisfies (1), then there is only one such measure and it is given by the formula in (2).
So both things yield the same thing: a description of the unique product measure.
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
$endgroup$
$begingroup$
And my assertion that $mathcal{A_{1}}times mathcal{A_{2}} subseteq mathcal{A_{1}}otimes mathcal{A_{2}}$ is correct, right?
$endgroup$
– SABOY
Jan 18 at 15:21
$begingroup$
Yes, of course. In fact, $mathcal{A_1} otimes mathcal{A_2}$ is the $sigma$-field generated by the "measurable rectangles" $mathcal{A_1} times mathcal{A}_2$
$endgroup$
– Math_QED
Jan 18 at 15:31
$begingroup$
But what other sets are there in $mathcal{A_{1}}otimes mathcal{A_{2}}$ that are not in $mathcal{A_{1}}times mathcal{A_{2}}$?
$endgroup$
– SABOY
Jan 20 at 15:03
$begingroup$
Do you know what the notation $sigma(mathcal{T})$ means (smallest sigma algebra generated by tau).
$endgroup$
– Math_QED
Jan 20 at 15:05
1
$begingroup$
It will depend on the sigma algebra's involved and a precise description will not be possible in general. But intuitively every set that you can make with (countably many) $cap, cup, ^c$ of the generating set will be in the sigma-algebra.
$endgroup$
– Math_QED
Jan 20 at 15:13
|
show 3 more comments
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
The actual product measure is given in (2).
If a measure on $mathcal{A_1} otimes mathcal{A_2}$ satisfies (1), then there is only one such measure and it is given by the formula in (2).
So both things yield the same thing: a description of the unique product measure.
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
$endgroup$
$begingroup$
And my assertion that $mathcal{A_{1}}times mathcal{A_{2}} subseteq mathcal{A_{1}}otimes mathcal{A_{2}}$ is correct, right?
$endgroup$
– SABOY
Jan 18 at 15:21
$begingroup$
Yes, of course. In fact, $mathcal{A_1} otimes mathcal{A_2}$ is the $sigma$-field generated by the "measurable rectangles" $mathcal{A_1} times mathcal{A}_2$
$endgroup$
– Math_QED
Jan 18 at 15:31
$begingroup$
But what other sets are there in $mathcal{A_{1}}otimes mathcal{A_{2}}$ that are not in $mathcal{A_{1}}times mathcal{A_{2}}$?
$endgroup$
– SABOY
Jan 20 at 15:03
$begingroup$
Do you know what the notation $sigma(mathcal{T})$ means (smallest sigma algebra generated by tau).
$endgroup$
– Math_QED
Jan 20 at 15:05
1
$begingroup$
It will depend on the sigma algebra's involved and a precise description will not be possible in general. But intuitively every set that you can make with (countably many) $cap, cup, ^c$ of the generating set will be in the sigma-algebra.
$endgroup$
– Math_QED
Jan 20 at 15:13
|
show 3 more comments
$begingroup$
The actual product measure is given in (2).
If a measure on $mathcal{A_1} otimes mathcal{A_2}$ satisfies (1), then there is only one such measure and it is given by the formula in (2).
So both things yield the same thing: a description of the unique product measure.
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
$endgroup$
$begingroup$
And my assertion that $mathcal{A_{1}}times mathcal{A_{2}} subseteq mathcal{A_{1}}otimes mathcal{A_{2}}$ is correct, right?
$endgroup$
– SABOY
Jan 18 at 15:21
$begingroup$
Yes, of course. In fact, $mathcal{A_1} otimes mathcal{A_2}$ is the $sigma$-field generated by the "measurable rectangles" $mathcal{A_1} times mathcal{A}_2$
$endgroup$
– Math_QED
Jan 18 at 15:31
$begingroup$
But what other sets are there in $mathcal{A_{1}}otimes mathcal{A_{2}}$ that are not in $mathcal{A_{1}}times mathcal{A_{2}}$?
$endgroup$
– SABOY
Jan 20 at 15:03
$begingroup$
Do you know what the notation $sigma(mathcal{T})$ means (smallest sigma algebra generated by tau).
$endgroup$
– Math_QED
Jan 20 at 15:05
1
$begingroup$
It will depend on the sigma algebra's involved and a precise description will not be possible in general. But intuitively every set that you can make with (countably many) $cap, cup, ^c$ of the generating set will be in the sigma-algebra.
$endgroup$
– Math_QED
Jan 20 at 15:13
|
show 3 more comments
$begingroup$
The actual product measure is given in (2).
If a measure on $mathcal{A_1} otimes mathcal{A_2}$ satisfies (1), then there is only one such measure and it is given by the formula in (2).
So both things yield the same thing: a description of the unique product measure.
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
$endgroup$
The actual product measure is given in (2).
If a measure on $mathcal{A_1} otimes mathcal{A_2}$ satisfies (1), then there is only one such measure and it is given by the formula in (2).
So both things yield the same thing: a description of the unique product measure.
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
answered Jan 18 at 15:14
Math_QEDMath_QED
7,64531452
7,64531452
$begingroup$
And my assertion that $mathcal{A_{1}}times mathcal{A_{2}} subseteq mathcal{A_{1}}otimes mathcal{A_{2}}$ is correct, right?
$endgroup$
– SABOY
Jan 18 at 15:21
$begingroup$
Yes, of course. In fact, $mathcal{A_1} otimes mathcal{A_2}$ is the $sigma$-field generated by the "measurable rectangles" $mathcal{A_1} times mathcal{A}_2$
$endgroup$
– Math_QED
Jan 18 at 15:31
$begingroup$
But what other sets are there in $mathcal{A_{1}}otimes mathcal{A_{2}}$ that are not in $mathcal{A_{1}}times mathcal{A_{2}}$?
$endgroup$
– SABOY
Jan 20 at 15:03
$begingroup$
Do you know what the notation $sigma(mathcal{T})$ means (smallest sigma algebra generated by tau).
$endgroup$
– Math_QED
Jan 20 at 15:05
1
$begingroup$
It will depend on the sigma algebra's involved and a precise description will not be possible in general. But intuitively every set that you can make with (countably many) $cap, cup, ^c$ of the generating set will be in the sigma-algebra.
$endgroup$
– Math_QED
Jan 20 at 15:13
|
show 3 more comments
$begingroup$
And my assertion that $mathcal{A_{1}}times mathcal{A_{2}} subseteq mathcal{A_{1}}otimes mathcal{A_{2}}$ is correct, right?
$endgroup$
– SABOY
Jan 18 at 15:21
$begingroup$
Yes, of course. In fact, $mathcal{A_1} otimes mathcal{A_2}$ is the $sigma$-field generated by the "measurable rectangles" $mathcal{A_1} times mathcal{A}_2$
$endgroup$
– Math_QED
Jan 18 at 15:31
$begingroup$
But what other sets are there in $mathcal{A_{1}}otimes mathcal{A_{2}}$ that are not in $mathcal{A_{1}}times mathcal{A_{2}}$?
$endgroup$
– SABOY
Jan 20 at 15:03
$begingroup$
Do you know what the notation $sigma(mathcal{T})$ means (smallest sigma algebra generated by tau).
$endgroup$
– Math_QED
Jan 20 at 15:05
1
$begingroup$
It will depend on the sigma algebra's involved and a precise description will not be possible in general. But intuitively every set that you can make with (countably many) $cap, cup, ^c$ of the generating set will be in the sigma-algebra.
$endgroup$
– Math_QED
Jan 20 at 15:13
$begingroup$
And my assertion that $mathcal{A_{1}}times mathcal{A_{2}} subseteq mathcal{A_{1}}otimes mathcal{A_{2}}$ is correct, right?
$endgroup$
– SABOY
Jan 18 at 15:21
$begingroup$
And my assertion that $mathcal{A_{1}}times mathcal{A_{2}} subseteq mathcal{A_{1}}otimes mathcal{A_{2}}$ is correct, right?
$endgroup$
– SABOY
Jan 18 at 15:21
$begingroup$
Yes, of course. In fact, $mathcal{A_1} otimes mathcal{A_2}$ is the $sigma$-field generated by the "measurable rectangles" $mathcal{A_1} times mathcal{A}_2$
$endgroup$
– Math_QED
Jan 18 at 15:31
$begingroup$
Yes, of course. In fact, $mathcal{A_1} otimes mathcal{A_2}$ is the $sigma$-field generated by the "measurable rectangles" $mathcal{A_1} times mathcal{A}_2$
$endgroup$
– Math_QED
Jan 18 at 15:31
$begingroup$
But what other sets are there in $mathcal{A_{1}}otimes mathcal{A_{2}}$ that are not in $mathcal{A_{1}}times mathcal{A_{2}}$?
$endgroup$
– SABOY
Jan 20 at 15:03
$begingroup$
But what other sets are there in $mathcal{A_{1}}otimes mathcal{A_{2}}$ that are not in $mathcal{A_{1}}times mathcal{A_{2}}$?
$endgroup$
– SABOY
Jan 20 at 15:03
$begingroup$
Do you know what the notation $sigma(mathcal{T})$ means (smallest sigma algebra generated by tau).
$endgroup$
– Math_QED
Jan 20 at 15:05
$begingroup$
Do you know what the notation $sigma(mathcal{T})$ means (smallest sigma algebra generated by tau).
$endgroup$
– Math_QED
Jan 20 at 15:05
1
1
$begingroup$
It will depend on the sigma algebra's involved and a precise description will not be possible in general. But intuitively every set that you can make with (countably many) $cap, cup, ^c$ of the generating set will be in the sigma-algebra.
$endgroup$
– Math_QED
Jan 20 at 15:13
$begingroup$
It will depend on the sigma algebra's involved and a precise description will not be possible in general. But intuitively every set that you can make with (countably many) $cap, cup, ^c$ of the generating set will be in the sigma-algebra.
$endgroup$
– Math_QED
Jan 20 at 15:13
|
show 3 more comments
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