Average (determine with Lebesgue integral/measure)
$begingroup$
Let $(X, mathcal{A}, mu)$ be a measure space and $A,B in mathcal{A}$ with $0<mu(A)<infty$.
How to compute the average of the indicator function $mathbf1_{B}$,
$⨍_A mathbf1_{B};dmu$ in terms of $mu$?
Since the definition of the average of an integral is $⨍_A f=⨍_A f dmu:=frac{1}{mu(A)} cdot int_A f d mu$ I tried:
$⨍_A mathbf1_{B};dmu = frac{1}{mu(A)} cdot int_A mathbf1_{B} d mu$.
Here I'm not sure how to continue as $mu(A)$ isn't given.
I thought about using:
$⨍_A mathbf1_{B};dmu = ⨍_X mathbf1_{B} cdot mathbf1_{B};dmu$, but it doesn't work.
measure-theory lebesgue-integral
asked Dec 15 '18 at 14:30
TartulopTartulop
706
$endgroup$
add a comment |
$begingroup$
Let $(X, mathcal{A}, mu)$ be a measure space and $A,B in mathcal{A}$ with $0<mu(A)<infty$.
How to compute the average of the indicator function $mathbf1_{B}$,
$⨍_A mathbf1_{B};dmu$ in terms of $mu$?
Since the definition of the average of an integral is $⨍_A f=⨍_A f dmu:=frac{1}{mu(A)} cdot int_A f d mu$ I tried:
$⨍_A mathbf1_{B};dmu = frac{1}{mu(A)} cdot int_A mathbf1_{B} d mu$.
Here I'm not sure how to continue as $mu(A)$ isn't given.
I thought about using:
$⨍_A mathbf1_{B};dmu = ⨍_X mathbf1_{B} cdot mathbf1_{B};dmu$, but it doesn't work.
measure-theory lebesgue-integral
asked Dec 15 '18 at 14:30
TartulopTartulop
706
$endgroup$
2
$begingroup$
It gives $frac{mu(Acap B)}{mu(A)}.$ You can't do better without more information
$endgroup$
– idm
Dec 15 '18 at 14:35
2
$begingroup$
$int_A mathbf1_{B} d mu=mu(Acap B)$
$endgroup$
– Matematleta
Dec 15 '18 at 14:36
add a comment |
$begingroup$
Let $(X, mathcal{A}, mu)$ be a measure space and $A,B in mathcal{A}$ with $0<mu(A)<infty$.
How to compute the average of the indicator function $mathbf1_{B}$,
$⨍_A mathbf1_{B};dmu$ in terms of $mu$?
Since the definition of the average of an integral is $⨍_A f=⨍_A f dmu:=frac{1}{mu(A)} cdot int_A f d mu$ I tried:
$⨍_A mathbf1_{B};dmu = frac{1}{mu(A)} cdot int_A mathbf1_{B} d mu$.
Here I'm not sure how to continue as $mu(A)$ isn't given.
I thought about using:
$⨍_A mathbf1_{B};dmu = ⨍_X mathbf1_{B} cdot mathbf1_{B};dmu$, but it doesn't work.
measure-theory lebesgue-integral
asked Dec 15 '18 at 14:30
TartulopTartulop
706
$endgroup$
Let $(X, mathcal{A}, mu)$ be a measure space and $A,B in mathcal{A}$ with $0<mu(A)<infty$.
How to compute the average of the indicator function $mathbf1_{B}$,
$⨍_A mathbf1_{B};dmu$ in terms of $mu$?
Since the definition of the average of an integral is $⨍_A f=⨍_A f dmu:=frac{1}{mu(A)} cdot int_A f d mu$ I tried:
$⨍_A mathbf1_{B};dmu = frac{1}{mu(A)} cdot int_A mathbf1_{B} d mu$.
Here I'm not sure how to continue as $mu(A)$ isn't given.
I thought about using:
$⨍_A mathbf1_{B};dmu = ⨍_X mathbf1_{B} cdot mathbf1_{B};dmu$, but it doesn't work.
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Dec 15 '18 at 14:30
TartulopTartulop
706
asked Dec 15 '18 at 14:30
TartulopTartulop
706
asked Dec 15 '18 at 14:30
TartulopTartulop
706
asked Dec 15 '18 at 14:30
TartulopTartulop
706
asked Dec 15 '18 at 14:30
TartulopTartulop
706
706
2
$begingroup$
It gives $frac{mu(Acap B)}{mu(A)}.$ You can't do better without more information
$endgroup$
– idm
Dec 15 '18 at 14:35
2
$begingroup$
$int_A mathbf1_{B} d mu=mu(Acap B)$
$endgroup$
– Matematleta
Dec 15 '18 at 14:36
add a comment |
2
$begingroup$
It gives $frac{mu(Acap B)}{mu(A)}.$ You can't do better without more information
$endgroup$
– idm
Dec 15 '18 at 14:35
2
$begingroup$
$int_A mathbf1_{B} d mu=mu(Acap B)$
$endgroup$
– Matematleta
Dec 15 '18 at 14:36
2
2
$begingroup$
It gives $frac{mu(Acap B)}{mu(A)}.$ You can't do better without more information
$endgroup$
– idm
Dec 15 '18 at 14:35
$begingroup$
It gives $frac{mu(Acap B)}{mu(A)}.$ You can't do better without more information
$endgroup$
– idm
Dec 15 '18 at 14:35
2
2
$begingroup$
$int_A mathbf1_{B} d mu=mu(Acap B)$
$endgroup$
– Matematleta
Dec 15 '18 at 14:36
$begingroup$
$int_A mathbf1_{B} d mu=mu(Acap B)$
$endgroup$
– Matematleta
Dec 15 '18 at 14:36
add a comment |
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2
$begingroup$
It gives $frac{mu(Acap B)}{mu(A)}.$ You can't do better without more information
$endgroup$
– idm
Dec 15 '18 at 14:35
2
$begingroup$
$int_A mathbf1_{B} d mu=mu(Acap B)$
$endgroup$
– Matematleta
Dec 15 '18 at 14:36