Measure on a sigma-algebra with integral
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
asked Nov 18 at 16:03
Tartulop
656
add a comment |
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
asked Nov 18 at 16:03
Tartulop
656
add a comment |
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
asked Nov 18 at 16:03
Tartulop
656
Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.
Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.
How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?
I tried it with:
$mu_f(emptyset)=int_emptyset f dmu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Nov 18 at 16:03
Tartulop
656
asked Nov 18 at 16:03
Tartulop
656
edited Nov 20 at 15:19
asked Nov 18 at 16:03
Tartulop
656
asked Nov 18 at 16:03
Tartulop
656
asked Nov 18 at 16:03
Tartulop
656
656
add a comment |
add a comment |
1 Answer
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$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 at 16:16
drhab
97.8k544129
add a comment |
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1 Answer
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1 Answer
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$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 at 16:16
drhab
97.8k544129
add a comment |
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 at 16:16
drhab
97.8k544129
add a comment |
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 at 16:16
drhab
97.8k544129
$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$
Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:
$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$
answered Nov 18 at 16:16
drhab
97.8k544129
answered Nov 18 at 16:16
drhab
97.8k544129
answered Nov 18 at 16:16
drhab
97.8k544129
answered Nov 18 at 16:16
drhab
97.8k544129
97.8k544129
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