Radon-Nikodym derivatives. Royden, Problem 33.
$begingroup$
Radon-Nikodym derivatives. Show that that Radon-Nikodym derivative $[frac{dnu}{dmu}]$ has the following properties:
a. If $null mu$ and $f$ is a nonnegative measurable function, then $int fdnu=int f[frac{dnu}{dmu}]dmu.$
Solution by Royden's solution: Let $(X,mathcal{B},mu)$ be a $sigma$-finite measure space and let $nu$ be a measure on $mathcal{B}$ which is absolutely continuous with respect to $mu$. Let $X=bigcup X_i$ with $mu(X_i)<infty$. We may assume the $X_i$ are pairwise disjoint.
For each $i$, let $mathcal{B}_i=left{Einmathcal{B}:Esubset X_iright}$, $mu_i=left.muright|_{mathcal{B_i}}$ and $nu_i=left.nuright|_{mathcal{B_i}}$. Then $(X_i,mathcal{B}_i, mu_i)$ is a finite measure space and $nu_i<<mu_i$. Thus for each $i$ there is a nonnegative $mu_i$-measurable function $f_i$ such that $nu_i(E)=int_E f_idmu_i$ forall $Ein mathcal{B}_i$. Define $f$ by $f(x)=f_i(x)$ if $xin X_i$.
If $Esubset X$, then $Ecap X_iin mathcal{B}_i$ for each $i$.
Thus $nu(E)=sum nu(Ecap X_i)=sum nu_i(Ecap X_i)=sum int_{Ecap X_i} f_idmu_i=sum int_{Ecap X_i} fdmu=int_E fdmu$.
I have doubts:
Why $mathcal{B}_i$ is a sigma-algebra? (If $Ein mathcal{B}_i$, then $Esubset X_i$ but ${X_i}^csubset E^c$. i.e. $E^cnotin mathcal{B}_i$...)
real-analysis measure-theory elementary-set-theory radon-nikodym
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
$endgroup$
add a comment |
$begingroup$
Radon-Nikodym derivatives. Show that that Radon-Nikodym derivative $[frac{dnu}{dmu}]$ has the following properties:
a. If $null mu$ and $f$ is a nonnegative measurable function, then $int fdnu=int f[frac{dnu}{dmu}]dmu.$
Solution by Royden's solution: Let $(X,mathcal{B},mu)$ be a $sigma$-finite measure space and let $nu$ be a measure on $mathcal{B}$ which is absolutely continuous with respect to $mu$. Let $X=bigcup X_i$ with $mu(X_i)<infty$. We may assume the $X_i$ are pairwise disjoint.
For each $i$, let $mathcal{B}_i=left{Einmathcal{B}:Esubset X_iright}$, $mu_i=left.muright|_{mathcal{B_i}}$ and $nu_i=left.nuright|_{mathcal{B_i}}$. Then $(X_i,mathcal{B}_i, mu_i)$ is a finite measure space and $nu_i<<mu_i$. Thus for each $i$ there is a nonnegative $mu_i$-measurable function $f_i$ such that $nu_i(E)=int_E f_idmu_i$ forall $Ein mathcal{B}_i$. Define $f$ by $f(x)=f_i(x)$ if $xin X_i$.
If $Esubset X$, then $Ecap X_iin mathcal{B}_i$ for each $i$.
Thus $nu(E)=sum nu(Ecap X_i)=sum nu_i(Ecap X_i)=sum int_{Ecap X_i} f_idmu_i=sum int_{Ecap X_i} fdmu=int_E fdmu$.
I have doubts:
Why $mathcal{B}_i$ is a sigma-algebra? (If $Ein mathcal{B}_i$, then $Esubset X_i$ but ${X_i}^csubset E^c$. i.e. $E^cnotin mathcal{B}_i$...)
real-analysis measure-theory elementary-set-theory radon-nikodym
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
$endgroup$
3
$begingroup$
The complement of $E$ is relative to $X_i$, not to $X$. That is to say, $cal B_i$ is a $sigma$-algebra of subsets of $X_i$.
$endgroup$
– Umberto P.
Jan 3 at 21:28
$begingroup$
Oh, true! Thank you
$endgroup$
– eraldcoil
Jan 3 at 21:31
add a comment |
$begingroup$
Radon-Nikodym derivatives. Show that that Radon-Nikodym derivative $[frac{dnu}{dmu}]$ has the following properties:
a. If $null mu$ and $f$ is a nonnegative measurable function, then $int fdnu=int f[frac{dnu}{dmu}]dmu.$
Solution by Royden's solution: Let $(X,mathcal{B},mu)$ be a $sigma$-finite measure space and let $nu$ be a measure on $mathcal{B}$ which is absolutely continuous with respect to $mu$. Let $X=bigcup X_i$ with $mu(X_i)<infty$. We may assume the $X_i$ are pairwise disjoint.
For each $i$, let $mathcal{B}_i=left{Einmathcal{B}:Esubset X_iright}$, $mu_i=left.muright|_{mathcal{B_i}}$ and $nu_i=left.nuright|_{mathcal{B_i}}$. Then $(X_i,mathcal{B}_i, mu_i)$ is a finite measure space and $nu_i<<mu_i$. Thus for each $i$ there is a nonnegative $mu_i$-measurable function $f_i$ such that $nu_i(E)=int_E f_idmu_i$ forall $Ein mathcal{B}_i$. Define $f$ by $f(x)=f_i(x)$ if $xin X_i$.
If $Esubset X$, then $Ecap X_iin mathcal{B}_i$ for each $i$.
Thus $nu(E)=sum nu(Ecap X_i)=sum nu_i(Ecap X_i)=sum int_{Ecap X_i} f_idmu_i=sum int_{Ecap X_i} fdmu=int_E fdmu$.
I have doubts:
Why $mathcal{B}_i$ is a sigma-algebra? (If $Ein mathcal{B}_i$, then $Esubset X_i$ but ${X_i}^csubset E^c$. i.e. $E^cnotin mathcal{B}_i$...)
real-analysis measure-theory elementary-set-theory radon-nikodym
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
$endgroup$
Radon-Nikodym derivatives. Show that that Radon-Nikodym derivative $[frac{dnu}{dmu}]$ has the following properties:
a. If $null mu$ and $f$ is a nonnegative measurable function, then $int fdnu=int f[frac{dnu}{dmu}]dmu.$
Solution by Royden's solution: Let $(X,mathcal{B},mu)$ be a $sigma$-finite measure space and let $nu$ be a measure on $mathcal{B}$ which is absolutely continuous with respect to $mu$. Let $X=bigcup X_i$ with $mu(X_i)<infty$. We may assume the $X_i$ are pairwise disjoint.
For each $i$, let $mathcal{B}_i=left{Einmathcal{B}:Esubset X_iright}$, $mu_i=left.muright|_{mathcal{B_i}}$ and $nu_i=left.nuright|_{mathcal{B_i}}$. Then $(X_i,mathcal{B}_i, mu_i)$ is a finite measure space and $nu_i<<mu_i$. Thus for each $i$ there is a nonnegative $mu_i$-measurable function $f_i$ such that $nu_i(E)=int_E f_idmu_i$ forall $Ein mathcal{B}_i$. Define $f$ by $f(x)=f_i(x)$ if $xin X_i$.
If $Esubset X$, then $Ecap X_iin mathcal{B}_i$ for each $i$.
Thus $nu(E)=sum nu(Ecap X_i)=sum nu_i(Ecap X_i)=sum int_{Ecap X_i} f_idmu_i=sum int_{Ecap X_i} fdmu=int_E fdmu$.
I have doubts:
Why $mathcal{B}_i$ is a sigma-algebra? (If $Ein mathcal{B}_i$, then $Esubset X_i$ but ${X_i}^csubset E^c$. i.e. $E^cnotin mathcal{B}_i$...)
real-analysis measure-theory elementary-set-theory radon-nikodym
real-analysis measure-theory elementary-set-theory radon-nikodym
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
edited Jan 3 at 23:05
Davide Giraudo
126k16150261
126k16150261
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
asked Jan 3 at 21:22
eraldcoileraldcoil
390211
390211
3
$begingroup$
The complement of $E$ is relative to $X_i$, not to $X$. That is to say, $cal B_i$ is a $sigma$-algebra of subsets of $X_i$.
$endgroup$
– Umberto P.
Jan 3 at 21:28
$begingroup$
Oh, true! Thank you
$endgroup$
– eraldcoil
Jan 3 at 21:31
add a comment |
3
$begingroup$
The complement of $E$ is relative to $X_i$, not to $X$. That is to say, $cal B_i$ is a $sigma$-algebra of subsets of $X_i$.
$endgroup$
– Umberto P.
Jan 3 at 21:28
$begingroup$
Oh, true! Thank you
$endgroup$
– eraldcoil
Jan 3 at 21:31
3
3
$begingroup$
The complement of $E$ is relative to $X_i$, not to $X$. That is to say, $cal B_i$ is a $sigma$-algebra of subsets of $X_i$.
$endgroup$
– Umberto P.
Jan 3 at 21:28
$begingroup$
The complement of $E$ is relative to $X_i$, not to $X$. That is to say, $cal B_i$ is a $sigma$-algebra of subsets of $X_i$.
$endgroup$
– Umberto P.
Jan 3 at 21:28
$begingroup$
Oh, true! Thank you
$endgroup$
– eraldcoil
Jan 3 at 21:31
$begingroup$
Oh, true! Thank you
$endgroup$
– eraldcoil
Jan 3 at 21:31
add a comment |
0
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$begingroup$
The complement of $E$ is relative to $X_i$, not to $X$. That is to say, $cal B_i$ is a $sigma$-algebra of subsets of $X_i$.
$endgroup$
– Umberto P.
Jan 3 at 21:28
$begingroup$
Oh, true! Thank you
$endgroup$
– eraldcoil
Jan 3 at 21:31