Confused on $ in mathcal{L}(X,mu)$ and $in mathcal{L}(X,nu)$
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I have come across an uncertainty in a proof I have just read:
Let $(X,mathcal{A})$ be measure space and $mu, nu, eta$ be $sigma-$finite measures such that $eta ll nu$ and $nu ll mu$
Show that $frac{deta}{dmu}=frac{deta}{dnu}frac{dnu}{dmu}, mu-$a.e.
My professor defines $(A_{n}^s)_{s}subseteq X$ s.t. $S(A_{n}^s)<infty, forall n in mathbb N$, whereby $A_{n}^{s}subseteq A_{n+1}^{s}$and $bigcup_{n in mathbb n}A_{n}^s=X, forall s in {mu, nu, eta}$ and $A_{n}:=A_{n}^{mu}cap A_{n}^{nu}cap A_{n}^{eta}$
Then following continuity from below, for all $A in mathcal{A}$
$eta(A)=lim_{nto infty}eta(A cap A_{n})=lim_{nto infty}int_{A cap A_{n}}frac{deta}{dnu}dnu=lim_{nto infty}int_{A}chi_{A_{n}}frac{deta}{dnu}{dnu}$
Problem: Then it states $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,mu)$ because $eta(A_{n})<infty$. Surely this is not correct. If anything it should be $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,nu)$ since by definition of the Radon-Nikodym derivative $frac{deta}{dnu}$ is measurable and $geq 0$. Furthermore, is the statement of this fact even necessary as $Acap A_{n}in mathcal{A}$ and is therefore by definition measurable w.r.t. $eta$.
I am only interested in the problem I laid out, not the RTP.
real-analysis measure-theory radon-nikodym
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
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add a comment |
$begingroup$
I have come across an uncertainty in a proof I have just read:
Let $(X,mathcal{A})$ be measure space and $mu, nu, eta$ be $sigma-$finite measures such that $eta ll nu$ and $nu ll mu$
Show that $frac{deta}{dmu}=frac{deta}{dnu}frac{dnu}{dmu}, mu-$a.e.
My professor defines $(A_{n}^s)_{s}subseteq X$ s.t. $S(A_{n}^s)<infty, forall n in mathbb N$, whereby $A_{n}^{s}subseteq A_{n+1}^{s}$and $bigcup_{n in mathbb n}A_{n}^s=X, forall s in {mu, nu, eta}$ and $A_{n}:=A_{n}^{mu}cap A_{n}^{nu}cap A_{n}^{eta}$
Then following continuity from below, for all $A in mathcal{A}$
$eta(A)=lim_{nto infty}eta(A cap A_{n})=lim_{nto infty}int_{A cap A_{n}}frac{deta}{dnu}dnu=lim_{nto infty}int_{A}chi_{A_{n}}frac{deta}{dnu}{dnu}$
Problem: Then it states $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,mu)$ because $eta(A_{n})<infty$. Surely this is not correct. If anything it should be $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,nu)$ since by definition of the Radon-Nikodym derivative $frac{deta}{dnu}$ is measurable and $geq 0$. Furthermore, is the statement of this fact even necessary as $Acap A_{n}in mathcal{A}$ and is therefore by definition measurable w.r.t. $eta$.
I am only interested in the problem I laid out, not the RTP.
real-analysis measure-theory radon-nikodym
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
$endgroup$
add a comment |
$begingroup$
I have come across an uncertainty in a proof I have just read:
Let $(X,mathcal{A})$ be measure space and $mu, nu, eta$ be $sigma-$finite measures such that $eta ll nu$ and $nu ll mu$
Show that $frac{deta}{dmu}=frac{deta}{dnu}frac{dnu}{dmu}, mu-$a.e.
My professor defines $(A_{n}^s)_{s}subseteq X$ s.t. $S(A_{n}^s)<infty, forall n in mathbb N$, whereby $A_{n}^{s}subseteq A_{n+1}^{s}$and $bigcup_{n in mathbb n}A_{n}^s=X, forall s in {mu, nu, eta}$ and $A_{n}:=A_{n}^{mu}cap A_{n}^{nu}cap A_{n}^{eta}$
Then following continuity from below, for all $A in mathcal{A}$
$eta(A)=lim_{nto infty}eta(A cap A_{n})=lim_{nto infty}int_{A cap A_{n}}frac{deta}{dnu}dnu=lim_{nto infty}int_{A}chi_{A_{n}}frac{deta}{dnu}{dnu}$
Problem: Then it states $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,mu)$ because $eta(A_{n})<infty$. Surely this is not correct. If anything it should be $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,nu)$ since by definition of the Radon-Nikodym derivative $frac{deta}{dnu}$ is measurable and $geq 0$. Furthermore, is the statement of this fact even necessary as $Acap A_{n}in mathcal{A}$ and is therefore by definition measurable w.r.t. $eta$.
I am only interested in the problem I laid out, not the RTP.
real-analysis measure-theory radon-nikodym
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
$endgroup$
I have come across an uncertainty in a proof I have just read:
Let $(X,mathcal{A})$ be measure space and $mu, nu, eta$ be $sigma-$finite measures such that $eta ll nu$ and $nu ll mu$
Show that $frac{deta}{dmu}=frac{deta}{dnu}frac{dnu}{dmu}, mu-$a.e.
My professor defines $(A_{n}^s)_{s}subseteq X$ s.t. $S(A_{n}^s)<infty, forall n in mathbb N$, whereby $A_{n}^{s}subseteq A_{n+1}^{s}$and $bigcup_{n in mathbb n}A_{n}^s=X, forall s in {mu, nu, eta}$ and $A_{n}:=A_{n}^{mu}cap A_{n}^{nu}cap A_{n}^{eta}$
Then following continuity from below, for all $A in mathcal{A}$
$eta(A)=lim_{nto infty}eta(A cap A_{n})=lim_{nto infty}int_{A cap A_{n}}frac{deta}{dnu}dnu=lim_{nto infty}int_{A}chi_{A_{n}}frac{deta}{dnu}{dnu}$
Problem: Then it states $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,mu)$ because $eta(A_{n})<infty$. Surely this is not correct. If anything it should be $chi_{A_{n}}frac{deta}{dnu}in mathcal{L}(X,nu)$ since by definition of the Radon-Nikodym derivative $frac{deta}{dnu}$ is measurable and $geq 0$. Furthermore, is the statement of this fact even necessary as $Acap A_{n}in mathcal{A}$ and is therefore by definition measurable w.r.t. $eta$.
I am only interested in the problem I laid out, not the RTP.
real-analysis measure-theory radon-nikodym
real-analysis measure-theory radon-nikodym
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
asked Dec 30 '18 at 22:04
SABOYSABOY
600311
600311
add a comment |
add a comment |
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