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

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 $\int g d\nu=\int g \frac{d\nu}{d\mu}d\mu$

b. We have $\nu << \lambda $ and $\frac{d\nu}{d\lambda}$=$ \frac{d\nu}{d\lambda}\frac{d\mu}{d\lambda} $ $\lambda$ a.e.

The proof in Folland is given as for part b:

I don't understand how g=$X_E \frac{d\nu}{d\mu}$ results in: $\nu(E)= \int_E \frac{d\nu}{d\mu}d\mu=\int_E \frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}d\lambda$? 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.

Answer 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 $$\int\limits g d\nu = \int\limits g f d\mu,$$ 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 $\nu(\mathrm{E}) = \int\limits_\mathrm{E} d\nu = \int\limits \mathbf{1}_\mathrm{E} f d\mu,$ but by the result of a., he also knows that $\int g d\mu = \int gh d\lambda,$ 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 $\nu(\mathrm{E}) = \int \mathbf{1}_\mathrm{E} fh d\lambda = \int\limits_\mathrm{E} fh d\lambda,$ hence, by the Lebesgue-Nikodym theorem, $fh$ is a density for $\nu$ with respect to $\lambda.$ Q.E.D.

Answer 2

The function $f:=d\nu/d\mu$ is the Radon-Nykodim derivative of $\nu$ w.r.t $\mu$. That is $$ \nu(E)=\int_E fd\mu=\int \chi_E\frac{d\nu}{d\mu}d\mu. $$ By part (a), setting $h:=\chi_Ef$ and letting $f':=d\mu/d\lambda$, we get $$ \int \chi_E\frac{d\nu}{d\mu}d\mu=\int hd\mu\overset{(a)}{=}\int hf'd\lambda. $$

Answer 3

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

a.) If $g\in L^1(\nu)$, then $g(d\nu/d\mu)\in L^1(\nu)$ and $$\int g d\nu = \int g \frac{d\nu}{d\mu}d\mu$$

b.) We have that $\nu\ll \lambda$, and $$\frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu}\frac{d\mu}{d\lambda} \ \lambda \ \text{a.e.}$$

Proof a.) Let us prove that \begin{equation}\label{e1}\int g d\nu = \int g \frac{d\nu}{d\mu}d\mu \end{equation}

First, note that for any $E \in M$, if $g = \chi_{E}$ then, by definition of $d\nu/d\mu$, we have that $$\int g d\nu = \nu(E)= \int_E\frac{d\nu}{d\mu}d\mu = \int \chi_{E}\frac{d\nu}{d\mu}d\mu =\int g\left(\frac{d\nu}{d\mu}\right)d\mu$$ So we have proved the case $g = \chi_{E}$.

Now, suppose $g$ is a simple function, that is $g=\sum_{i=1}^n\chi_{E_i}$. Then we have, using what we have just proved, $$\int g d\nu = \int \sum_{i=1}^n\chi_{E_i} d\nu =\sum_{i=1}^n\int \chi_{E_i} d\nu =\sum_{i=1}^n\int \chi_{E_i} \left(\frac{d\nu}{d\mu}\right)d\mu =\int\left(\sum_{i=1}^n \chi_{E_i} \right)\left(\frac{d\nu}{d\mu}\right)d\mu =\int g \left(\frac{d\nu}{d\mu}\right)d\mu$$

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 $g_n\to f$. Then we have $g_n\left(\frac{d\nu}{d\mu}\right)^+$ converges monotonically to $f\left(\frac{d\nu}{d\mu}\right)^+$ and $g_n\left(\frac{d\nu}{d\mu}\right)^-$ converges monotonically to $f\left(\frac{d\nu}{d\mu}\right)^-$. So, using the result for simple functions and the Monotone Convergence Theorem (actually three times) we have \begin{align*}\int f d\nu &= \lim_{n \to \infty} \int g_n d\nu= \lim_{n \to \infty} \int g_n \left(\frac{d\nu}{d\mu}\right)d\mu= \\ & =\lim_{n \to \infty} \int g_n \left(\frac{d\nu}{d\mu}\right)^+d\mu-\lim_{n \to \infty} \int g_n \left(\frac{d\nu}{d\mu}\right)^-d\mu= \\ & = \int f \left(\frac{d\nu}{d\mu}\right)^+d\mu- \int f \left(\frac{d\nu}{d\mu}\right)^-d\mu= \\ & = \int f \left(\frac{d\nu}{d\mu}\right)d\mu \end{align*}

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

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

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

$$\int g d\mu = \int g \left(\frac{d\mu}{d\lambda}\right)d\lambda$$

Given any set $E\in M$, we know that $\chi_E \left(\frac{d\nu}{d\mu}\right)^+$ and $\chi_E \left(\frac{d\nu}{d\mu}\right)^-$ are non-negative measurable functions, so we have \begin{align*}\nu(E)&= \int \chi_E \left(\frac{d\nu}{d\mu}\right) d\mu =\\ &=\int \chi_E \left(\frac{d\nu}{d\mu}\right)^+ d\mu-\int \chi_E \left(\frac{d\nu}{d\mu}\right)^- d\mu=\\ &=\int \chi_E \left(\frac{d\nu}{d\mu}\right)^+ \left(\frac{d\mu}{d\lambda}\right)d\lambda-\int \chi_E \left(\frac{d\nu}{d\mu}\right)^- \left(\frac{d\mu}{d\lambda}\right)d\lambda=\\ &=\int \chi_E \left ( \left(\frac{d\nu}{d\mu}\right)^+ -\left(\frac{d\nu}{d\mu}\right)^-\right) \left(\frac{d\mu}{d\lambda}\right)d\lambda=\\ &=\int \chi_E \left(\frac{d\nu}{d\mu}\right) \left(\frac{d\mu}{d\lambda}\right)d\lambda\\ &=\int_E \left(\frac{d\nu}{d\mu}\right) \left(\frac{d\mu}{d\lambda}\right)d\lambda \end{align*} So, from above and using proposition 2.23, we have $$\left(\frac{d\nu}{d\lambda}\right)=\left(\frac{d\nu}{d\mu}\right) \left(\frac{d\mu}{d\lambda}\right)$$