Measurable functions (Integral)
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Let $sigma_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
$sigma_x(F):=begin{cases}1,&text{if }x in Ftext{ }\0,&text{otherwise } end{cases}quad$
How to prove that all maps $g:mathbb{R} to mathbb{R}$ are integrable with respect to $sigma_x$?
I know that $int_mathbb{R}g=g(x)$.
I tried to use that:
$int_mathbb{R}|g| dsigma_x= |g(x)|<infty$ so all $g$ are integrable with respect to $sigma_x$
But I'm not sure if this works.
measure-theory proof-verification lebesgue-integral
edited Nov 24 at 12:02
Scientifica
6,34141333
asked Nov 24 at 9:59
Nekarts
234
add a comment |
up vote
2
down vote
favorite
Let $sigma_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
$sigma_x(F):=begin{cases}1,&text{if }x in Ftext{ }\0,&text{otherwise } end{cases}quad$
How to prove that all maps $g:mathbb{R} to mathbb{R}$ are integrable with respect to $sigma_x$?
I know that $int_mathbb{R}g=g(x)$.
I tried to use that:
$int_mathbb{R}|g| dsigma_x= |g(x)|<infty$ so all $g$ are integrable with respect to $sigma_x$
But I'm not sure if this works.
measure-theory proof-verification lebesgue-integral
edited Nov 24 at 12:02
Scientifica
6,34141333
asked Nov 24 at 9:59
Nekarts
234
1
Your approach is correct.
– drhab
Nov 24 at 11:01
1
You may also want to state that all $mathbb{R}^n to mathbb{R}$ functions are measurable because all subsets of $mathbb{R}^n$ are $sigma_x$-measurable, so we don't need to worry about measurability of functions at all.
– Alex Vong
Nov 24 at 11:06
For a rigorous proof starting with definition of integral you have to first verify the equation $int g dsigma_x=g(x)$ for simple functions the take limits. Measurability is no issue because you are taking power set as the sigma algebra.
– Kavi Rama Murthy
Nov 24 at 11:53
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $sigma_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
$sigma_x(F):=begin{cases}1,&text{if }x in Ftext{ }\0,&text{otherwise } end{cases}quad$
How to prove that all maps $g:mathbb{R} to mathbb{R}$ are integrable with respect to $sigma_x$?
I know that $int_mathbb{R}g=g(x)$.
I tried to use that:
$int_mathbb{R}|g| dsigma_x= |g(x)|<infty$ so all $g$ are integrable with respect to $sigma_x$
But I'm not sure if this works.
measure-theory proof-verification lebesgue-integral
edited Nov 24 at 12:02
Scientifica
6,34141333
asked Nov 24 at 9:59
Nekarts
234
Let $sigma_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
$sigma_x(F):=begin{cases}1,&text{if }x in Ftext{ }\0,&text{otherwise } end{cases}quad$
How to prove that all maps $g:mathbb{R} to mathbb{R}$ are integrable with respect to $sigma_x$?
I know that $int_mathbb{R}g=g(x)$.
I tried to use that:
$int_mathbb{R}|g| dsigma_x= |g(x)|<infty$ so all $g$ are integrable with respect to $sigma_x$
But I'm not sure if this works.
measure-theory proof-verification lebesgue-integral
measure-theory proof-verification lebesgue-integral
edited Nov 24 at 12:02
Scientifica
6,34141333
asked Nov 24 at 9:59
Nekarts
234
edited Nov 24 at 12:02
Scientifica
6,34141333
asked Nov 24 at 9:59
Nekarts
234
edited Nov 24 at 12:02
Scientifica
6,34141333
edited Nov 24 at 12:02
Scientifica
6,34141333
edited Nov 24 at 12:02
Scientifica
6,34141333
6,34141333
asked Nov 24 at 9:59
Nekarts
234
asked Nov 24 at 9:59
Nekarts
234
asked Nov 24 at 9:59
Nekarts
234
234
1
Your approach is correct.
– drhab
Nov 24 at 11:01
1
You may also want to state that all $mathbb{R}^n to mathbb{R}$ functions are measurable because all subsets of $mathbb{R}^n$ are $sigma_x$-measurable, so we don't need to worry about measurability of functions at all.
– Alex Vong
Nov 24 at 11:06
For a rigorous proof starting with definition of integral you have to first verify the equation $int g dsigma_x=g(x)$ for simple functions the take limits. Measurability is no issue because you are taking power set as the sigma algebra.
– Kavi Rama Murthy
Nov 24 at 11:53
add a comment |
1
Your approach is correct.
– drhab
Nov 24 at 11:01
1
You may also want to state that all $mathbb{R}^n to mathbb{R}$ functions are measurable because all subsets of $mathbb{R}^n$ are $sigma_x$-measurable, so we don't need to worry about measurability of functions at all.
– Alex Vong
Nov 24 at 11:06
For a rigorous proof starting with definition of integral you have to first verify the equation $int g dsigma_x=g(x)$ for simple functions the take limits. Measurability is no issue because you are taking power set as the sigma algebra.
– Kavi Rama Murthy
Nov 24 at 11:53
1
1
Your approach is correct.
– drhab
Nov 24 at 11:01
Your approach is correct.
– drhab
Nov 24 at 11:01
1
1
You may also want to state that all $mathbb{R}^n to mathbb{R}$ functions are measurable because all subsets of $mathbb{R}^n$ are $sigma_x$-measurable, so we don't need to worry about measurability of functions at all.
– Alex Vong
Nov 24 at 11:06
You may also want to state that all $mathbb{R}^n to mathbb{R}$ functions are measurable because all subsets of $mathbb{R}^n$ are $sigma_x$-measurable, so we don't need to worry about measurability of functions at all.
– Alex Vong
Nov 24 at 11:06
For a rigorous proof starting with definition of integral you have to first verify the equation $int g dsigma_x=g(x)$ for simple functions the take limits. Measurability is no issue because you are taking power set as the sigma algebra.
– Kavi Rama Murthy
Nov 24 at 11:53
For a rigorous proof starting with definition of integral you have to first verify the equation $int g dsigma_x=g(x)$ for simple functions the take limits. Measurability is no issue because you are taking power set as the sigma algebra.
– Kavi Rama Murthy
Nov 24 at 11:53
add a comment |
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1
Your approach is correct.
– drhab
Nov 24 at 11:01
1
You may also want to state that all $mathbb{R}^n to mathbb{R}$ functions are measurable because all subsets of $mathbb{R}^n$ are $sigma_x$-measurable, so we don't need to worry about measurability of functions at all.
– Alex Vong
Nov 24 at 11:06
For a rigorous proof starting with definition of integral you have to first verify the equation $int g dsigma_x=g(x)$ for simple functions the take limits. Measurability is no issue because you are taking power set as the sigma algebra.
– Kavi Rama Murthy
Nov 24 at 11:53