Ytukyg

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:

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.

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.

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
$$
int chi_Efrac{dnu}{dmu}dmu=int hdmuoverset{(a)}{=}int hf'dlambda.
$$

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
$$left(frac{dnu}{dlambda}right)=left(frac{dnu}{dmu}right) left(frac{dmu}{dlambda}right)$$