Definition of map preserve the measure : why is it $mu(T^-1(A))=mu(A)$ and not $mu(T(A))=mu(A)$.Problem with the definition of Lebesgue Measure and borel sets!When is the composition operator assigned to a measure-preserving map unitary?Definition of doubling measureDoes changing the values of a $mathbb P$-a.s. defined map at countably many points preserve the map?Do differentiable functions preserve measure zero sets? Measurable sets?Why do we usually construct the Lebesgue Measure on finite outer measure sets before arbitrary measurable sets?Why do we need to declare a probability measure for the definition of stochastic processes?Why the probability distribution of a uniform random variable is the Lebesgue measure?Making sense of measure-theoretic definition of random variableIs the integral of a continuous map always a Radon measure?
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Definition of map preserve the measure : why is it $mu(T^-1(A))=mu(A)$ and not $mu(T(A))=mu(A)$.
Problem with the definition of Lebesgue Measure and borel sets!When is the composition operator assigned to a measure-preserving map unitary?Definition of doubling measureDoes changing the values of a $mathbb P$-a.s. defined map at countably many points preserve the map?Do differentiable functions preserve measure zero sets? Measurable sets?Why do we usually construct the Lebesgue Measure on finite outer measure sets before arbitrary measurable sets?Why do we need to declare a probability measure for the definition of stochastic processes?Why the probability distribution of a uniform random variable is the Lebesgue measure?Making sense of measure-theoretic definition of random variableIs the integral of a continuous map always a Radon measure?
$begingroup$
Let $(X, mathcal B,mathbb P)$ a probability space. In wikipedia, they say that a map $T:Xto X$ preserve the measure if $$mathbb P(T^-1(A))=mu(A),$$
for all $Ain mathcal B$.
Why a definition as $mathbb P(T(A))=mathbb P(A)$ would not work ? It looks more natural, no ?
measure-theory
New contributor
$endgroup$
add a comment |
$begingroup$
Let $(X, mathcal B,mathbb P)$ a probability space. In wikipedia, they say that a map $T:Xto X$ preserve the measure if $$mathbb P(T^-1(A))=mu(A),$$
for all $Ain mathcal B$.
Why a definition as $mathbb P(T(A))=mathbb P(A)$ would not work ? It looks more natural, no ?
measure-theory
New contributor
$endgroup$
add a comment |
$begingroup$
Let $(X, mathcal B,mathbb P)$ a probability space. In wikipedia, they say that a map $T:Xto X$ preserve the measure if $$mathbb P(T^-1(A))=mu(A),$$
for all $Ain mathcal B$.
Why a definition as $mathbb P(T(A))=mathbb P(A)$ would not work ? It looks more natural, no ?
measure-theory
New contributor
$endgroup$
Let $(X, mathcal B,mathbb P)$ a probability space. In wikipedia, they say that a map $T:Xto X$ preserve the measure if $$mathbb P(T^-1(A))=mu(A),$$
for all $Ain mathcal B$.
Why a definition as $mathbb P(T(A))=mathbb P(A)$ would not work ? It looks more natural, no ?
measure-theory
measure-theory
New contributor
New contributor
New contributor
asked Mar 11 at 10:09
PierrePierre
618
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$begingroup$
The image of a measurabe set under a measurable transformation may not be measurable but the inverse image is always measurable (by definition of measurability).
If $T$ is bijective and $T^-1$ is also measurable thee the two conditions are identical.
$endgroup$
add a comment |
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$begingroup$
The image of a measurabe set under a measurable transformation may not be measurable but the inverse image is always measurable (by definition of measurability).
If $T$ is bijective and $T^-1$ is also measurable thee the two conditions are identical.
$endgroup$
add a comment |
$begingroup$
The image of a measurabe set under a measurable transformation may not be measurable but the inverse image is always measurable (by definition of measurability).
If $T$ is bijective and $T^-1$ is also measurable thee the two conditions are identical.
$endgroup$
add a comment |
$begingroup$
The image of a measurabe set under a measurable transformation may not be measurable but the inverse image is always measurable (by definition of measurability).
If $T$ is bijective and $T^-1$ is also measurable thee the two conditions are identical.
$endgroup$
The image of a measurabe set under a measurable transformation may not be measurable but the inverse image is always measurable (by definition of measurability).
If $T$ is bijective and $T^-1$ is also measurable thee the two conditions are identical.
answered Mar 11 at 10:12
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
67.2k53067
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