Understanding Proposition 3.9 Folland Part 2 (Radon-Nikodym Derivative)












0














Suppose that $nu$ is a $sigma$-finite signed measure and $mu$, $lambda$ are $sigma$-finite measures on $(X,M)$ such that $nu<<mu$ and $mu << lambda$.



a. If g $in L_1(mu)$ then
$int g dnu=int g frac{dnu}{dmu}dmu$



b. We have $nu << lambda $ and $frac{dnu}{dlambda}$=$ frac{dnu}{dlambda}frac{dmu}{dlambda} $ $lambda$ a.e.



The proof in Folland is given as for part b:
enter image description here



I don't understand how g=$X_E frac{dnu}{dmu}$ results in:
$nu(E)= int_E frac{dnu}{dmu}dmu=int_E frac{dnu}{dmu}frac{dmu}{dlambda}dlambda$?
I know it sort of follows from part a but I don't exactly see all the steps. Any help would be much appreciated as I am struggling very deeply. Thank you.










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  • any help would be much appreciated struggling very much
    – kemb
    Dec 4 '18 at 3:44
















0














Suppose that $nu$ is a $sigma$-finite signed measure and $mu$, $lambda$ are $sigma$-finite measures on $(X,M)$ such that $nu<<mu$ and $mu << lambda$.



a. If g $in L_1(mu)$ then
$int g dnu=int g frac{dnu}{dmu}dmu$



b. We have $nu << lambda $ and $frac{dnu}{dlambda}$=$ frac{dnu}{dlambda}frac{dmu}{dlambda} $ $lambda$ a.e.



The proof in Folland is given as for part b:
enter image description here



I don't understand how g=$X_E frac{dnu}{dmu}$ results in:
$nu(E)= int_E frac{dnu}{dmu}dmu=int_E frac{dnu}{dmu}frac{dmu}{dlambda}dlambda$?
I know it sort of follows from part a but I don't exactly see all the steps. Any help would be much appreciated as I am struggling very deeply. Thank you.










share|cite|improve this question
























  • any help would be much appreciated struggling very much
    – kemb
    Dec 4 '18 at 3:44














0












0








0


1





Suppose that $nu$ is a $sigma$-finite signed measure and $mu$, $lambda$ are $sigma$-finite measures on $(X,M)$ such that $nu<<mu$ and $mu << lambda$.



a. If g $in L_1(mu)$ then
$int g dnu=int g frac{dnu}{dmu}dmu$



b. We have $nu << lambda $ and $frac{dnu}{dlambda}$=$ frac{dnu}{dlambda}frac{dmu}{dlambda} $ $lambda$ a.e.



The proof in Folland is given as for part b:
enter image description here



I don't understand how g=$X_E frac{dnu}{dmu}$ results in:
$nu(E)= int_E frac{dnu}{dmu}dmu=int_E frac{dnu}{dmu}frac{dmu}{dlambda}dlambda$?
I know it sort of follows from part a but I don't exactly see all the steps. Any help would be much appreciated as I am struggling very deeply. Thank you.










share|cite|improve this question















Suppose that $nu$ is a $sigma$-finite signed measure and $mu$, $lambda$ are $sigma$-finite measures on $(X,M)$ such that $nu<<mu$ and $mu << lambda$.



a. If g $in L_1(mu)$ then
$int g dnu=int g frac{dnu}{dmu}dmu$



b. We have $nu << lambda $ and $frac{dnu}{dlambda}$=$ frac{dnu}{dlambda}frac{dmu}{dlambda} $ $lambda$ a.e.



The proof in Folland is given as for part b:
enter image description here



I don't understand how g=$X_E frac{dnu}{dmu}$ results in:
$nu(E)= int_E frac{dnu}{dmu}dmu=int_E frac{dnu}{dmu}frac{dmu}{dlambda}dlambda$?
I know it sort of follows from part a but I don't exactly see all the steps. Any help would be much appreciated as I am struggling very deeply. Thank you.







real-analysis analysis probability-theory measure-theory radon-nikodym






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edited Dec 4 '18 at 1:37

























asked Dec 4 '18 at 1:26









kemb

683213




683213












  • any help would be much appreciated struggling very much
    – kemb
    Dec 4 '18 at 3:44


















  • any help would be much appreciated struggling very much
    – kemb
    Dec 4 '18 at 3:44
















any help would be much appreciated struggling very much
– kemb
Dec 4 '18 at 3:44




any help would be much appreciated struggling very much
– kemb
Dec 4 '18 at 3:44










3 Answers
3






active

oldest

votes


















1














I have never ever read Folland, so I am guessing what he is doing in a. He must define the set $mathscr{H}_+$ of the positive measurable functions $g$ such that
$$intlimits g dnu = intlimits g f dmu,$$
where $f$ is the corresponding density of $nu$ with respect to $mu.$ By the usual way Folland proves, I guess, that $mathscr{H}_+$ is all positive measurable functions hence the result of a. comes just by appealing to integrability.



As for b. he observes first that $nu(mathrm{E}) = intlimits_mathrm{E} dnu = intlimits mathbf{1}_mathrm{E} f dmu,$ but by the result of a., he also knows that $int g dmu = int gh dlambda,$ where $h$ is a density of $mu$ with respect to $lambda$ and $g$ is any integrable function, so applying this to $g = mathbf{1}_mathrm{E} f$ he concludes that $nu(mathrm{E}) = int mathbf{1}_mathrm{E} fh dlambda = intlimits_mathrm{E} fh dlambda,$ hence, by the Lebesgue-Nikodym theorem, $fh$ is a density for $nu$ with respect to $lambda.$ Q.E.D.






share|cite|improve this answer





















  • sorry I'm not familar with the term density. what does it mean. thanks
    – kemb
    Dec 4 '18 at 6:23










  • "Density" is borrowed from probability theory and it just means "Lebesgue-Nikodym derivative."
    – Will M.
    Dec 4 '18 at 6:24



















1














The function $f:=dnu/dmu$ is the Radon-Nykodim derivative of $nu$ w.r.t $mu$. That is
$$
nu(E)=int_E fdmu=int chi_Efrac{dnu}{dmu}dmu.
$$

By part (a), setting $h:=chi_Ef$ and letting $f':=dmu/dlambda$, we get
$$
int chi_Efrac{dnu}{dmu}dmu=int hdmuoverset{(a)}{=}int hf'dlambda.
$$






share|cite|improve this answer





















  • what is meant by $f'=dnu/dlambda$? Radon-Nikodym derivative; now we replace $nu$ and $mu$ by $mu$ and $lambda$? thanks
    – kemb
    Dec 4 '18 at 7:07












  • @kemb That's correct. Folland writes $dnu=f'dlambda$.
    – d.k.o.
    Dec 4 '18 at 7:22












  • I don't get how $int h d mu$ = $hf' dlambda$ by part a , could you explain this part?
    – kemb
    Dec 4 '18 at 8:04












  • Also why does Folland 'replace' $nu$ and $mu$ by $mu$ and $lambda$? Is it because now $nu << lambda$
    – kemb
    Dec 4 '18 at 8:06












  • @kemb Replace $g$ and $f$ is (a) by $h$ and $f'$.
    – d.k.o.
    Dec 4 '18 at 8:48



















0















Proposition 3.9:
Suppose that $nu$ is a $sigma$-finite measure and $lambda$ are $sigma$-finite measures on $(X,M)$ such that $null mu$ and $mull lambda$.



a.) If $gin L^1(nu)$, then $g(dnu/dmu)in L^1(nu)$ and $$int g dnu = int g frac{dnu}{dmu}dmu$$



b.) We have that $null lambda$, and $$frac{dnu}{dlambda} = frac{dnu}{dmu}frac{dmu}{dlambda} lambda text{a.e.}$$




Proof a.) Let us prove that
begin{equation}label{e1}int g dnu = int g frac{dnu}{dmu}dmu end{equation}



First, note that for any $E in M$, if $g = chi_{E}$ then, by definition of $dnu/dmu$, we have that
$$int g dnu = nu(E)= int_Efrac{dnu}{dmu}dmu = int chi_{E}frac{dnu}{dmu}dmu =int gleft(frac{dnu}{dmu}right)dmu$$
So we have proved the case $g = chi_{E}$.



Now, suppose $g$ is a simple function, that is $g=sum_{i=1}^nchi_{E_i}$. Then we have, using what we have just proved,
$$int g dnu = int sum_{i=1}^nchi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} left(frac{dnu}{dmu}right)dmu =intleft(sum_{i=1}^n chi_{E_i} right)left(frac{dnu}{dmu}right)dmu =int g left(frac{dnu}{dmu}right)dmu$$



So we have proved the case $g$ is a simple function.



Now suppose that $f$ is non-negative measurable function. Then there is ${g_n}_n$ is a monotone non-decreasing sequence of non-negative simples functions such that $g_nto f$. Then we have $g_nleft(frac{dnu}{dmu}right)^+$ converges monotonically to $fleft(frac{dnu}{dmu}right)^+$ and $g_nleft(frac{dnu}{dmu}right)^-$ converges monotonically to $fleft(frac{dnu}{dmu}right)^-$. So, using the result for simple functions and the Monotone Convergence Theorem (actually three times) we have
begin{align*}int f dnu &= lim_{n to infty} int g_n dnu= lim_{n to infty} int g_n left(frac{dnu}{dmu}right)dmu= \
& =lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^+dmu-lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^-dmu= \
& = int f left(frac{dnu}{dmu}right)^+dmu- int f left(frac{dnu}{dmu}right)^-dmu= \
& = int f left(frac{dnu}{dmu}right)dmu
end{align*}



Finally, let $gin L^1(nu)$. Then $g^+$ and $g^-$ are non-negative measurable functions. So from what we have just proved we have
$$int g dnu = int g^+ dnu-int g^- dnu=int g^+ left(frac{dnu}{dmu}right)dmu-int g^- left(frac{dnu}{dmu}right)dmu= int (g^+-g^-) left(frac{dnu}{dmu}right)dmu=int g left(frac{dnu}{dmu}right)dmu$$



Proof b.) For any measurable set $E$, if $lambda(E)=0$ then $mu(E)=0$ and then $nu(E)=0$. So we have that $nulllambda$.
Then, for any measurable set $E$,
begin{equation}label{e2}
nu(E) = int_E left(frac{dnu}{dlambda}right)dlambda
end{equation}



Now, from a.) for the measures $mu$ and $lambda$, we know that for every non-negative measurable function $g$ we have



$$int g dmu = int g left(frac{dmu}{dlambda}right)dlambda$$



Given any set $Ein M$, we know that $chi_E left(frac{dnu}{dmu}right)^+$ and $chi_E left(frac{dnu}{dmu}right)^-$ are non-negative measurable functions, so we have
begin{align*}nu(E)&= int chi_E left(frac{dnu}{dmu}right) dmu =\
&=int chi_E left(frac{dnu}{dmu}right)^+ dmu-int chi_E left(frac{dnu}{dmu}right)^- dmu=\
&=int chi_E left(frac{dnu}{dmu}right)^+ left(frac{dmu}{dlambda}right)dlambda-int chi_E left(frac{dnu}{dmu}right)^- left(frac{dmu}{dlambda}right)dlambda=\
&=int chi_E left ( left(frac{dnu}{dmu}right)^+ -left(frac{dnu}{dmu}right)^-right) left(frac{dmu}{dlambda}right)dlambda=\
&=int chi_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda\
&=int_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda
end{align*}

So, from above and using proposition 2.23, we have
$$left(frac{dnu}{dlambda}right)=left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)$$






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    3 Answers
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    3 Answers
    3






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    active

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    active

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    1














    I have never ever read Folland, so I am guessing what he is doing in a. He must define the set $mathscr{H}_+$ of the positive measurable functions $g$ such that
    $$intlimits g dnu = intlimits g f dmu,$$
    where $f$ is the corresponding density of $nu$ with respect to $mu.$ By the usual way Folland proves, I guess, that $mathscr{H}_+$ is all positive measurable functions hence the result of a. comes just by appealing to integrability.



    As for b. he observes first that $nu(mathrm{E}) = intlimits_mathrm{E} dnu = intlimits mathbf{1}_mathrm{E} f dmu,$ but by the result of a., he also knows that $int g dmu = int gh dlambda,$ where $h$ is a density of $mu$ with respect to $lambda$ and $g$ is any integrable function, so applying this to $g = mathbf{1}_mathrm{E} f$ he concludes that $nu(mathrm{E}) = int mathbf{1}_mathrm{E} fh dlambda = intlimits_mathrm{E} fh dlambda,$ hence, by the Lebesgue-Nikodym theorem, $fh$ is a density for $nu$ with respect to $lambda.$ Q.E.D.






    share|cite|improve this answer





















    • sorry I'm not familar with the term density. what does it mean. thanks
      – kemb
      Dec 4 '18 at 6:23










    • "Density" is borrowed from probability theory and it just means "Lebesgue-Nikodym derivative."
      – Will M.
      Dec 4 '18 at 6:24
















    1














    I have never ever read Folland, so I am guessing what he is doing in a. He must define the set $mathscr{H}_+$ of the positive measurable functions $g$ such that
    $$intlimits g dnu = intlimits g f dmu,$$
    where $f$ is the corresponding density of $nu$ with respect to $mu.$ By the usual way Folland proves, I guess, that $mathscr{H}_+$ is all positive measurable functions hence the result of a. comes just by appealing to integrability.



    As for b. he observes first that $nu(mathrm{E}) = intlimits_mathrm{E} dnu = intlimits mathbf{1}_mathrm{E} f dmu,$ but by the result of a., he also knows that $int g dmu = int gh dlambda,$ where $h$ is a density of $mu$ with respect to $lambda$ and $g$ is any integrable function, so applying this to $g = mathbf{1}_mathrm{E} f$ he concludes that $nu(mathrm{E}) = int mathbf{1}_mathrm{E} fh dlambda = intlimits_mathrm{E} fh dlambda,$ hence, by the Lebesgue-Nikodym theorem, $fh$ is a density for $nu$ with respect to $lambda.$ Q.E.D.






    share|cite|improve this answer





















    • sorry I'm not familar with the term density. what does it mean. thanks
      – kemb
      Dec 4 '18 at 6:23










    • "Density" is borrowed from probability theory and it just means "Lebesgue-Nikodym derivative."
      – Will M.
      Dec 4 '18 at 6:24














    1












    1








    1






    I have never ever read Folland, so I am guessing what he is doing in a. He must define the set $mathscr{H}_+$ of the positive measurable functions $g$ such that
    $$intlimits g dnu = intlimits g f dmu,$$
    where $f$ is the corresponding density of $nu$ with respect to $mu.$ By the usual way Folland proves, I guess, that $mathscr{H}_+$ is all positive measurable functions hence the result of a. comes just by appealing to integrability.



    As for b. he observes first that $nu(mathrm{E}) = intlimits_mathrm{E} dnu = intlimits mathbf{1}_mathrm{E} f dmu,$ but by the result of a., he also knows that $int g dmu = int gh dlambda,$ where $h$ is a density of $mu$ with respect to $lambda$ and $g$ is any integrable function, so applying this to $g = mathbf{1}_mathrm{E} f$ he concludes that $nu(mathrm{E}) = int mathbf{1}_mathrm{E} fh dlambda = intlimits_mathrm{E} fh dlambda,$ hence, by the Lebesgue-Nikodym theorem, $fh$ is a density for $nu$ with respect to $lambda.$ Q.E.D.






    share|cite|improve this answer












    I have never ever read Folland, so I am guessing what he is doing in a. He must define the set $mathscr{H}_+$ of the positive measurable functions $g$ such that
    $$intlimits g dnu = intlimits g f dmu,$$
    where $f$ is the corresponding density of $nu$ with respect to $mu.$ By the usual way Folland proves, I guess, that $mathscr{H}_+$ is all positive measurable functions hence the result of a. comes just by appealing to integrability.



    As for b. he observes first that $nu(mathrm{E}) = intlimits_mathrm{E} dnu = intlimits mathbf{1}_mathrm{E} f dmu,$ but by the result of a., he also knows that $int g dmu = int gh dlambda,$ where $h$ is a density of $mu$ with respect to $lambda$ and $g$ is any integrable function, so applying this to $g = mathbf{1}_mathrm{E} f$ he concludes that $nu(mathrm{E}) = int mathbf{1}_mathrm{E} fh dlambda = intlimits_mathrm{E} fh dlambda,$ hence, by the Lebesgue-Nikodym theorem, $fh$ is a density for $nu$ with respect to $lambda.$ Q.E.D.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 4 '18 at 6:14









    Will M.

    2,442314




    2,442314












    • sorry I'm not familar with the term density. what does it mean. thanks
      – kemb
      Dec 4 '18 at 6:23










    • "Density" is borrowed from probability theory and it just means "Lebesgue-Nikodym derivative."
      – Will M.
      Dec 4 '18 at 6:24


















    • sorry I'm not familar with the term density. what does it mean. thanks
      – kemb
      Dec 4 '18 at 6:23










    • "Density" is borrowed from probability theory and it just means "Lebesgue-Nikodym derivative."
      – Will M.
      Dec 4 '18 at 6:24
















    sorry I'm not familar with the term density. what does it mean. thanks
    – kemb
    Dec 4 '18 at 6:23




    sorry I'm not familar with the term density. what does it mean. thanks
    – kemb
    Dec 4 '18 at 6:23












    "Density" is borrowed from probability theory and it just means "Lebesgue-Nikodym derivative."
    – Will M.
    Dec 4 '18 at 6:24




    "Density" is borrowed from probability theory and it just means "Lebesgue-Nikodym derivative."
    – Will M.
    Dec 4 '18 at 6:24











    1














    The function $f:=dnu/dmu$ is the Radon-Nykodim derivative of $nu$ w.r.t $mu$. That is
    $$
    nu(E)=int_E fdmu=int chi_Efrac{dnu}{dmu}dmu.
    $$

    By part (a), setting $h:=chi_Ef$ and letting $f':=dmu/dlambda$, we get
    $$
    int chi_Efrac{dnu}{dmu}dmu=int hdmuoverset{(a)}{=}int hf'dlambda.
    $$






    share|cite|improve this answer





















    • what is meant by $f'=dnu/dlambda$? Radon-Nikodym derivative; now we replace $nu$ and $mu$ by $mu$ and $lambda$? thanks
      – kemb
      Dec 4 '18 at 7:07












    • @kemb That's correct. Folland writes $dnu=f'dlambda$.
      – d.k.o.
      Dec 4 '18 at 7:22












    • I don't get how $int h d mu$ = $hf' dlambda$ by part a , could you explain this part?
      – kemb
      Dec 4 '18 at 8:04












    • Also why does Folland 'replace' $nu$ and $mu$ by $mu$ and $lambda$? Is it because now $nu << lambda$
      – kemb
      Dec 4 '18 at 8:06












    • @kemb Replace $g$ and $f$ is (a) by $h$ and $f'$.
      – d.k.o.
      Dec 4 '18 at 8:48
















    1














    The function $f:=dnu/dmu$ is the Radon-Nykodim derivative of $nu$ w.r.t $mu$. That is
    $$
    nu(E)=int_E fdmu=int chi_Efrac{dnu}{dmu}dmu.
    $$

    By part (a), setting $h:=chi_Ef$ and letting $f':=dmu/dlambda$, we get
    $$
    int chi_Efrac{dnu}{dmu}dmu=int hdmuoverset{(a)}{=}int hf'dlambda.
    $$






    share|cite|improve this answer





















    • what is meant by $f'=dnu/dlambda$? Radon-Nikodym derivative; now we replace $nu$ and $mu$ by $mu$ and $lambda$? thanks
      – kemb
      Dec 4 '18 at 7:07












    • @kemb That's correct. Folland writes $dnu=f'dlambda$.
      – d.k.o.
      Dec 4 '18 at 7:22












    • I don't get how $int h d mu$ = $hf' dlambda$ by part a , could you explain this part?
      – kemb
      Dec 4 '18 at 8:04












    • Also why does Folland 'replace' $nu$ and $mu$ by $mu$ and $lambda$? Is it because now $nu << lambda$
      – kemb
      Dec 4 '18 at 8:06












    • @kemb Replace $g$ and $f$ is (a) by $h$ and $f'$.
      – d.k.o.
      Dec 4 '18 at 8:48














    1












    1








    1






    The function $f:=dnu/dmu$ is the Radon-Nykodim derivative of $nu$ w.r.t $mu$. That is
    $$
    nu(E)=int_E fdmu=int chi_Efrac{dnu}{dmu}dmu.
    $$

    By part (a), setting $h:=chi_Ef$ and letting $f':=dmu/dlambda$, we get
    $$
    int chi_Efrac{dnu}{dmu}dmu=int hdmuoverset{(a)}{=}int hf'dlambda.
    $$






    share|cite|improve this answer












    The function $f:=dnu/dmu$ is the Radon-Nykodim derivative of $nu$ w.r.t $mu$. That is
    $$
    nu(E)=int_E fdmu=int chi_Efrac{dnu}{dmu}dmu.
    $$

    By part (a), setting $h:=chi_Ef$ and letting $f':=dmu/dlambda$, we get
    $$
    int chi_Efrac{dnu}{dmu}dmu=int hdmuoverset{(a)}{=}int hf'dlambda.
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 4 '18 at 6:33









    d.k.o.

    8,612528




    8,612528












    • what is meant by $f'=dnu/dlambda$? Radon-Nikodym derivative; now we replace $nu$ and $mu$ by $mu$ and $lambda$? thanks
      – kemb
      Dec 4 '18 at 7:07












    • @kemb That's correct. Folland writes $dnu=f'dlambda$.
      – d.k.o.
      Dec 4 '18 at 7:22












    • I don't get how $int h d mu$ = $hf' dlambda$ by part a , could you explain this part?
      – kemb
      Dec 4 '18 at 8:04












    • Also why does Folland 'replace' $nu$ and $mu$ by $mu$ and $lambda$? Is it because now $nu << lambda$
      – kemb
      Dec 4 '18 at 8:06












    • @kemb Replace $g$ and $f$ is (a) by $h$ and $f'$.
      – d.k.o.
      Dec 4 '18 at 8:48


















    • what is meant by $f'=dnu/dlambda$? Radon-Nikodym derivative; now we replace $nu$ and $mu$ by $mu$ and $lambda$? thanks
      – kemb
      Dec 4 '18 at 7:07












    • @kemb That's correct. Folland writes $dnu=f'dlambda$.
      – d.k.o.
      Dec 4 '18 at 7:22












    • I don't get how $int h d mu$ = $hf' dlambda$ by part a , could you explain this part?
      – kemb
      Dec 4 '18 at 8:04












    • Also why does Folland 'replace' $nu$ and $mu$ by $mu$ and $lambda$? Is it because now $nu << lambda$
      – kemb
      Dec 4 '18 at 8:06












    • @kemb Replace $g$ and $f$ is (a) by $h$ and $f'$.
      – d.k.o.
      Dec 4 '18 at 8:48
















    what is meant by $f'=dnu/dlambda$? Radon-Nikodym derivative; now we replace $nu$ and $mu$ by $mu$ and $lambda$? thanks
    – kemb
    Dec 4 '18 at 7:07






    what is meant by $f'=dnu/dlambda$? Radon-Nikodym derivative; now we replace $nu$ and $mu$ by $mu$ and $lambda$? thanks
    – kemb
    Dec 4 '18 at 7:07














    @kemb That's correct. Folland writes $dnu=f'dlambda$.
    – d.k.o.
    Dec 4 '18 at 7:22






    @kemb That's correct. Folland writes $dnu=f'dlambda$.
    – d.k.o.
    Dec 4 '18 at 7:22














    I don't get how $int h d mu$ = $hf' dlambda$ by part a , could you explain this part?
    – kemb
    Dec 4 '18 at 8:04






    I don't get how $int h d mu$ = $hf' dlambda$ by part a , could you explain this part?
    – kemb
    Dec 4 '18 at 8:04














    Also why does Folland 'replace' $nu$ and $mu$ by $mu$ and $lambda$? Is it because now $nu << lambda$
    – kemb
    Dec 4 '18 at 8:06






    Also why does Folland 'replace' $nu$ and $mu$ by $mu$ and $lambda$? Is it because now $nu << lambda$
    – kemb
    Dec 4 '18 at 8:06














    @kemb Replace $g$ and $f$ is (a) by $h$ and $f'$.
    – d.k.o.
    Dec 4 '18 at 8:48




    @kemb Replace $g$ and $f$ is (a) by $h$ and $f'$.
    – d.k.o.
    Dec 4 '18 at 8:48











    0















    Proposition 3.9:
    Suppose that $nu$ is a $sigma$-finite measure and $lambda$ are $sigma$-finite measures on $(X,M)$ such that $null mu$ and $mull lambda$.



    a.) If $gin L^1(nu)$, then $g(dnu/dmu)in L^1(nu)$ and $$int g dnu = int g frac{dnu}{dmu}dmu$$



    b.) We have that $null lambda$, and $$frac{dnu}{dlambda} = frac{dnu}{dmu}frac{dmu}{dlambda} lambda text{a.e.}$$




    Proof a.) Let us prove that
    begin{equation}label{e1}int g dnu = int g frac{dnu}{dmu}dmu end{equation}



    First, note that for any $E in M$, if $g = chi_{E}$ then, by definition of $dnu/dmu$, we have that
    $$int g dnu = nu(E)= int_Efrac{dnu}{dmu}dmu = int chi_{E}frac{dnu}{dmu}dmu =int gleft(frac{dnu}{dmu}right)dmu$$
    So we have proved the case $g = chi_{E}$.



    Now, suppose $g$ is a simple function, that is $g=sum_{i=1}^nchi_{E_i}$. Then we have, using what we have just proved,
    $$int g dnu = int sum_{i=1}^nchi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} left(frac{dnu}{dmu}right)dmu =intleft(sum_{i=1}^n chi_{E_i} right)left(frac{dnu}{dmu}right)dmu =int g left(frac{dnu}{dmu}right)dmu$$



    So we have proved the case $g$ is a simple function.



    Now suppose that $f$ is non-negative measurable function. Then there is ${g_n}_n$ is a monotone non-decreasing sequence of non-negative simples functions such that $g_nto f$. Then we have $g_nleft(frac{dnu}{dmu}right)^+$ converges monotonically to $fleft(frac{dnu}{dmu}right)^+$ and $g_nleft(frac{dnu}{dmu}right)^-$ converges monotonically to $fleft(frac{dnu}{dmu}right)^-$. So, using the result for simple functions and the Monotone Convergence Theorem (actually three times) we have
    begin{align*}int f dnu &= lim_{n to infty} int g_n dnu= lim_{n to infty} int g_n left(frac{dnu}{dmu}right)dmu= \
    & =lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^+dmu-lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^-dmu= \
    & = int f left(frac{dnu}{dmu}right)^+dmu- int f left(frac{dnu}{dmu}right)^-dmu= \
    & = int f left(frac{dnu}{dmu}right)dmu
    end{align*}



    Finally, let $gin L^1(nu)$. Then $g^+$ and $g^-$ are non-negative measurable functions. So from what we have just proved we have
    $$int g dnu = int g^+ dnu-int g^- dnu=int g^+ left(frac{dnu}{dmu}right)dmu-int g^- left(frac{dnu}{dmu}right)dmu= int (g^+-g^-) left(frac{dnu}{dmu}right)dmu=int g left(frac{dnu}{dmu}right)dmu$$



    Proof b.) For any measurable set $E$, if $lambda(E)=0$ then $mu(E)=0$ and then $nu(E)=0$. So we have that $nulllambda$.
    Then, for any measurable set $E$,
    begin{equation}label{e2}
    nu(E) = int_E left(frac{dnu}{dlambda}right)dlambda
    end{equation}



    Now, from a.) for the measures $mu$ and $lambda$, we know that for every non-negative measurable function $g$ we have



    $$int g dmu = int g left(frac{dmu}{dlambda}right)dlambda$$



    Given any set $Ein M$, we know that $chi_E left(frac{dnu}{dmu}right)^+$ and $chi_E left(frac{dnu}{dmu}right)^-$ are non-negative measurable functions, so we have
    begin{align*}nu(E)&= int chi_E left(frac{dnu}{dmu}right) dmu =\
    &=int chi_E left(frac{dnu}{dmu}right)^+ dmu-int chi_E left(frac{dnu}{dmu}right)^- dmu=\
    &=int chi_E left(frac{dnu}{dmu}right)^+ left(frac{dmu}{dlambda}right)dlambda-int chi_E left(frac{dnu}{dmu}right)^- left(frac{dmu}{dlambda}right)dlambda=\
    &=int chi_E left ( left(frac{dnu}{dmu}right)^+ -left(frac{dnu}{dmu}right)^-right) left(frac{dmu}{dlambda}right)dlambda=\
    &=int chi_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda\
    &=int_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda
    end{align*}

    So, from above and using proposition 2.23, we have
    $$left(frac{dnu}{dlambda}right)=left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)$$






    share|cite|improve this answer


























      0















      Proposition 3.9:
      Suppose that $nu$ is a $sigma$-finite measure and $lambda$ are $sigma$-finite measures on $(X,M)$ such that $null mu$ and $mull lambda$.



      a.) If $gin L^1(nu)$, then $g(dnu/dmu)in L^1(nu)$ and $$int g dnu = int g frac{dnu}{dmu}dmu$$



      b.) We have that $null lambda$, and $$frac{dnu}{dlambda} = frac{dnu}{dmu}frac{dmu}{dlambda} lambda text{a.e.}$$




      Proof a.) Let us prove that
      begin{equation}label{e1}int g dnu = int g frac{dnu}{dmu}dmu end{equation}



      First, note that for any $E in M$, if $g = chi_{E}$ then, by definition of $dnu/dmu$, we have that
      $$int g dnu = nu(E)= int_Efrac{dnu}{dmu}dmu = int chi_{E}frac{dnu}{dmu}dmu =int gleft(frac{dnu}{dmu}right)dmu$$
      So we have proved the case $g = chi_{E}$.



      Now, suppose $g$ is a simple function, that is $g=sum_{i=1}^nchi_{E_i}$. Then we have, using what we have just proved,
      $$int g dnu = int sum_{i=1}^nchi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} left(frac{dnu}{dmu}right)dmu =intleft(sum_{i=1}^n chi_{E_i} right)left(frac{dnu}{dmu}right)dmu =int g left(frac{dnu}{dmu}right)dmu$$



      So we have proved the case $g$ is a simple function.



      Now suppose that $f$ is non-negative measurable function. Then there is ${g_n}_n$ is a monotone non-decreasing sequence of non-negative simples functions such that $g_nto f$. Then we have $g_nleft(frac{dnu}{dmu}right)^+$ converges monotonically to $fleft(frac{dnu}{dmu}right)^+$ and $g_nleft(frac{dnu}{dmu}right)^-$ converges monotonically to $fleft(frac{dnu}{dmu}right)^-$. So, using the result for simple functions and the Monotone Convergence Theorem (actually three times) we have
      begin{align*}int f dnu &= lim_{n to infty} int g_n dnu= lim_{n to infty} int g_n left(frac{dnu}{dmu}right)dmu= \
      & =lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^+dmu-lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^-dmu= \
      & = int f left(frac{dnu}{dmu}right)^+dmu- int f left(frac{dnu}{dmu}right)^-dmu= \
      & = int f left(frac{dnu}{dmu}right)dmu
      end{align*}



      Finally, let $gin L^1(nu)$. Then $g^+$ and $g^-$ are non-negative measurable functions. So from what we have just proved we have
      $$int g dnu = int g^+ dnu-int g^- dnu=int g^+ left(frac{dnu}{dmu}right)dmu-int g^- left(frac{dnu}{dmu}right)dmu= int (g^+-g^-) left(frac{dnu}{dmu}right)dmu=int g left(frac{dnu}{dmu}right)dmu$$



      Proof b.) For any measurable set $E$, if $lambda(E)=0$ then $mu(E)=0$ and then $nu(E)=0$. So we have that $nulllambda$.
      Then, for any measurable set $E$,
      begin{equation}label{e2}
      nu(E) = int_E left(frac{dnu}{dlambda}right)dlambda
      end{equation}



      Now, from a.) for the measures $mu$ and $lambda$, we know that for every non-negative measurable function $g$ we have



      $$int g dmu = int g left(frac{dmu}{dlambda}right)dlambda$$



      Given any set $Ein M$, we know that $chi_E left(frac{dnu}{dmu}right)^+$ and $chi_E left(frac{dnu}{dmu}right)^-$ are non-negative measurable functions, so we have
      begin{align*}nu(E)&= int chi_E left(frac{dnu}{dmu}right) dmu =\
      &=int chi_E left(frac{dnu}{dmu}right)^+ dmu-int chi_E left(frac{dnu}{dmu}right)^- dmu=\
      &=int chi_E left(frac{dnu}{dmu}right)^+ left(frac{dmu}{dlambda}right)dlambda-int chi_E left(frac{dnu}{dmu}right)^- left(frac{dmu}{dlambda}right)dlambda=\
      &=int chi_E left ( left(frac{dnu}{dmu}right)^+ -left(frac{dnu}{dmu}right)^-right) left(frac{dmu}{dlambda}right)dlambda=\
      &=int chi_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda\
      &=int_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda
      end{align*}

      So, from above and using proposition 2.23, we have
      $$left(frac{dnu}{dlambda}right)=left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)$$






      share|cite|improve this answer
























        0












        0








        0







        Proposition 3.9:
        Suppose that $nu$ is a $sigma$-finite measure and $lambda$ are $sigma$-finite measures on $(X,M)$ such that $null mu$ and $mull lambda$.



        a.) If $gin L^1(nu)$, then $g(dnu/dmu)in L^1(nu)$ and $$int g dnu = int g frac{dnu}{dmu}dmu$$



        b.) We have that $null lambda$, and $$frac{dnu}{dlambda} = frac{dnu}{dmu}frac{dmu}{dlambda} lambda text{a.e.}$$




        Proof a.) Let us prove that
        begin{equation}label{e1}int g dnu = int g frac{dnu}{dmu}dmu end{equation}



        First, note that for any $E in M$, if $g = chi_{E}$ then, by definition of $dnu/dmu$, we have that
        $$int g dnu = nu(E)= int_Efrac{dnu}{dmu}dmu = int chi_{E}frac{dnu}{dmu}dmu =int gleft(frac{dnu}{dmu}right)dmu$$
        So we have proved the case $g = chi_{E}$.



        Now, suppose $g$ is a simple function, that is $g=sum_{i=1}^nchi_{E_i}$. Then we have, using what we have just proved,
        $$int g dnu = int sum_{i=1}^nchi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} left(frac{dnu}{dmu}right)dmu =intleft(sum_{i=1}^n chi_{E_i} right)left(frac{dnu}{dmu}right)dmu =int g left(frac{dnu}{dmu}right)dmu$$



        So we have proved the case $g$ is a simple function.



        Now suppose that $f$ is non-negative measurable function. Then there is ${g_n}_n$ is a monotone non-decreasing sequence of non-negative simples functions such that $g_nto f$. Then we have $g_nleft(frac{dnu}{dmu}right)^+$ converges monotonically to $fleft(frac{dnu}{dmu}right)^+$ and $g_nleft(frac{dnu}{dmu}right)^-$ converges monotonically to $fleft(frac{dnu}{dmu}right)^-$. So, using the result for simple functions and the Monotone Convergence Theorem (actually three times) we have
        begin{align*}int f dnu &= lim_{n to infty} int g_n dnu= lim_{n to infty} int g_n left(frac{dnu}{dmu}right)dmu= \
        & =lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^+dmu-lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^-dmu= \
        & = int f left(frac{dnu}{dmu}right)^+dmu- int f left(frac{dnu}{dmu}right)^-dmu= \
        & = int f left(frac{dnu}{dmu}right)dmu
        end{align*}



        Finally, let $gin L^1(nu)$. Then $g^+$ and $g^-$ are non-negative measurable functions. So from what we have just proved we have
        $$int g dnu = int g^+ dnu-int g^- dnu=int g^+ left(frac{dnu}{dmu}right)dmu-int g^- left(frac{dnu}{dmu}right)dmu= int (g^+-g^-) left(frac{dnu}{dmu}right)dmu=int g left(frac{dnu}{dmu}right)dmu$$



        Proof b.) For any measurable set $E$, if $lambda(E)=0$ then $mu(E)=0$ and then $nu(E)=0$. So we have that $nulllambda$.
        Then, for any measurable set $E$,
        begin{equation}label{e2}
        nu(E) = int_E left(frac{dnu}{dlambda}right)dlambda
        end{equation}



        Now, from a.) for the measures $mu$ and $lambda$, we know that for every non-negative measurable function $g$ we have



        $$int g dmu = int g left(frac{dmu}{dlambda}right)dlambda$$



        Given any set $Ein M$, we know that $chi_E left(frac{dnu}{dmu}right)^+$ and $chi_E left(frac{dnu}{dmu}right)^-$ are non-negative measurable functions, so we have
        begin{align*}nu(E)&= int chi_E left(frac{dnu}{dmu}right) dmu =\
        &=int chi_E left(frac{dnu}{dmu}right)^+ dmu-int chi_E left(frac{dnu}{dmu}right)^- dmu=\
        &=int chi_E left(frac{dnu}{dmu}right)^+ left(frac{dmu}{dlambda}right)dlambda-int chi_E left(frac{dnu}{dmu}right)^- left(frac{dmu}{dlambda}right)dlambda=\
        &=int chi_E left ( left(frac{dnu}{dmu}right)^+ -left(frac{dnu}{dmu}right)^-right) left(frac{dmu}{dlambda}right)dlambda=\
        &=int chi_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda\
        &=int_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda
        end{align*}

        So, from above and using proposition 2.23, we have
        $$left(frac{dnu}{dlambda}right)=left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)$$






        share|cite|improve this answer













        Proposition 3.9:
        Suppose that $nu$ is a $sigma$-finite measure and $lambda$ are $sigma$-finite measures on $(X,M)$ such that $null mu$ and $mull lambda$.



        a.) If $gin L^1(nu)$, then $g(dnu/dmu)in L^1(nu)$ and $$int g dnu = int g frac{dnu}{dmu}dmu$$



        b.) We have that $null lambda$, and $$frac{dnu}{dlambda} = frac{dnu}{dmu}frac{dmu}{dlambda} lambda text{a.e.}$$




        Proof a.) Let us prove that
        begin{equation}label{e1}int g dnu = int g frac{dnu}{dmu}dmu end{equation}



        First, note that for any $E in M$, if $g = chi_{E}$ then, by definition of $dnu/dmu$, we have that
        $$int g dnu = nu(E)= int_Efrac{dnu}{dmu}dmu = int chi_{E}frac{dnu}{dmu}dmu =int gleft(frac{dnu}{dmu}right)dmu$$
        So we have proved the case $g = chi_{E}$.



        Now, suppose $g$ is a simple function, that is $g=sum_{i=1}^nchi_{E_i}$. Then we have, using what we have just proved,
        $$int g dnu = int sum_{i=1}^nchi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} dnu =sum_{i=1}^nint chi_{E_i} left(frac{dnu}{dmu}right)dmu =intleft(sum_{i=1}^n chi_{E_i} right)left(frac{dnu}{dmu}right)dmu =int g left(frac{dnu}{dmu}right)dmu$$



        So we have proved the case $g$ is a simple function.



        Now suppose that $f$ is non-negative measurable function. Then there is ${g_n}_n$ is a monotone non-decreasing sequence of non-negative simples functions such that $g_nto f$. Then we have $g_nleft(frac{dnu}{dmu}right)^+$ converges monotonically to $fleft(frac{dnu}{dmu}right)^+$ and $g_nleft(frac{dnu}{dmu}right)^-$ converges monotonically to $fleft(frac{dnu}{dmu}right)^-$. So, using the result for simple functions and the Monotone Convergence Theorem (actually three times) we have
        begin{align*}int f dnu &= lim_{n to infty} int g_n dnu= lim_{n to infty} int g_n left(frac{dnu}{dmu}right)dmu= \
        & =lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^+dmu-lim_{n to infty} int g_n left(frac{dnu}{dmu}right)^-dmu= \
        & = int f left(frac{dnu}{dmu}right)^+dmu- int f left(frac{dnu}{dmu}right)^-dmu= \
        & = int f left(frac{dnu}{dmu}right)dmu
        end{align*}



        Finally, let $gin L^1(nu)$. Then $g^+$ and $g^-$ are non-negative measurable functions. So from what we have just proved we have
        $$int g dnu = int g^+ dnu-int g^- dnu=int g^+ left(frac{dnu}{dmu}right)dmu-int g^- left(frac{dnu}{dmu}right)dmu= int (g^+-g^-) left(frac{dnu}{dmu}right)dmu=int g left(frac{dnu}{dmu}right)dmu$$



        Proof b.) For any measurable set $E$, if $lambda(E)=0$ then $mu(E)=0$ and then $nu(E)=0$. So we have that $nulllambda$.
        Then, for any measurable set $E$,
        begin{equation}label{e2}
        nu(E) = int_E left(frac{dnu}{dlambda}right)dlambda
        end{equation}



        Now, from a.) for the measures $mu$ and $lambda$, we know that for every non-negative measurable function $g$ we have



        $$int g dmu = int g left(frac{dmu}{dlambda}right)dlambda$$



        Given any set $Ein M$, we know that $chi_E left(frac{dnu}{dmu}right)^+$ and $chi_E left(frac{dnu}{dmu}right)^-$ are non-negative measurable functions, so we have
        begin{align*}nu(E)&= int chi_E left(frac{dnu}{dmu}right) dmu =\
        &=int chi_E left(frac{dnu}{dmu}right)^+ dmu-int chi_E left(frac{dnu}{dmu}right)^- dmu=\
        &=int chi_E left(frac{dnu}{dmu}right)^+ left(frac{dmu}{dlambda}right)dlambda-int chi_E left(frac{dnu}{dmu}right)^- left(frac{dmu}{dlambda}right)dlambda=\
        &=int chi_E left ( left(frac{dnu}{dmu}right)^+ -left(frac{dnu}{dmu}right)^-right) left(frac{dmu}{dlambda}right)dlambda=\
        &=int chi_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda\
        &=int_E left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)dlambda
        end{align*}

        So, from above and using proposition 2.23, we have
        $$left(frac{dnu}{dlambda}right)=left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 6:34









        Wolfy

        2,31311038




        2,31311038






























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