Measure theory: on absolutely continuous measures
up vote
1
down vote
favorite
Let F be a random variable defined on a probability space $(Omega,mathcal F, mathbb P)$ with values on $(E,mathcal E, lambda)$, where $E=(-1,1)$ and $lambda$ is the Lebesgue measure.
To prove that the law of $F$ is absolutely continuous (with respect to the Lebesgue measure on $mathbb R$) I know that I should prove that for every measurable set $Ainmathcal E$ such that $lambda(A)=0$, we have $mathbb Pcirc F^{-1}(A)=0.$
It that equivalent to show that for any measurable function $g:(-1,1)rightarrow [0,1]$ such that $int_{-1}^1 g(x)dx=0$, we have $mathbb E[g(F)]=0$ ? Why?
probability-theory measure-theory lebesgue-measure absolute-continuity
asked Nov 23 at 15:33
claudia
64
add a comment |
up vote
1
down vote
favorite
Let F be a random variable defined on a probability space $(Omega,mathcal F, mathbb P)$ with values on $(E,mathcal E, lambda)$, where $E=(-1,1)$ and $lambda$ is the Lebesgue measure.
To prove that the law of $F$ is absolutely continuous (with respect to the Lebesgue measure on $mathbb R$) I know that I should prove that for every measurable set $Ainmathcal E$ such that $lambda(A)=0$, we have $mathbb Pcirc F^{-1}(A)=0.$
It that equivalent to show that for any measurable function $g:(-1,1)rightarrow [0,1]$ such that $int_{-1}^1 g(x)dx=0$, we have $mathbb E[g(F)]=0$ ? Why?
probability-theory measure-theory lebesgue-measure absolute-continuity
asked Nov 23 at 15:33
claudia
64
Radon–Nikodym theorem might help
– PSG
Nov 23 at 18:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let F be a random variable defined on a probability space $(Omega,mathcal F, mathbb P)$ with values on $(E,mathcal E, lambda)$, where $E=(-1,1)$ and $lambda$ is the Lebesgue measure.
To prove that the law of $F$ is absolutely continuous (with respect to the Lebesgue measure on $mathbb R$) I know that I should prove that for every measurable set $Ainmathcal E$ such that $lambda(A)=0$, we have $mathbb Pcirc F^{-1}(A)=0.$
It that equivalent to show that for any measurable function $g:(-1,1)rightarrow [0,1]$ such that $int_{-1}^1 g(x)dx=0$, we have $mathbb E[g(F)]=0$ ? Why?
probability-theory measure-theory lebesgue-measure absolute-continuity
asked Nov 23 at 15:33
claudia
64
Let F be a random variable defined on a probability space $(Omega,mathcal F, mathbb P)$ with values on $(E,mathcal E, lambda)$, where $E=(-1,1)$ and $lambda$ is the Lebesgue measure.
To prove that the law of $F$ is absolutely continuous (with respect to the Lebesgue measure on $mathbb R$) I know that I should prove that for every measurable set $Ainmathcal E$ such that $lambda(A)=0$, we have $mathbb Pcirc F^{-1}(A)=0.$
It that equivalent to show that for any measurable function $g:(-1,1)rightarrow [0,1]$ such that $int_{-1}^1 g(x)dx=0$, we have $mathbb E[g(F)]=0$ ? Why?
probability-theory measure-theory lebesgue-measure absolute-continuity
probability-theory measure-theory lebesgue-measure absolute-continuity
asked Nov 23 at 15:33
claudia
64
asked Nov 23 at 15:33
claudia
64
edited Nov 23 at 17:56
asked Nov 23 at 15:33
claudia
64
asked Nov 23 at 15:33
claudia
64
asked Nov 23 at 15:33
claudia
64
64
Radon–Nikodym theorem might help
– PSG
Nov 23 at 18:13
add a comment |
Radon–Nikodym theorem might help
– PSG
Nov 23 at 18:13
Radon–Nikodym theorem might help
– PSG
Nov 23 at 18:13
Radon–Nikodym theorem might help
– PSG
Nov 23 at 18:13
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010485%2fmeasure-theory-on-absolutely-continuous-measures%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Radon–Nikodym theorem might help
– PSG
Nov 23 at 18:13