Lebesgue Radon Nikodym Theorem
Let $\nu$ be a $\sigma$-finite signed measure on $(X,\mathcal{A})$ and $\mu$ a $\sigma$-finite positive measure on $(X,\mathcal{A})$
- There exist unique $\sigma$-finite signed measures $\rho,\lambda$ on $(X,\mathcal{A})$ such that $$\nu=\rho+\lambda\qquad \rho \ll\mu,\qquad\lambda \perp \mu.$$
- There exists an extended $\mu$-integrable function $f$ such that $d\rho=f\,d\mu$ i.e.$$\nu=f\,d\mu+\lambda$$
- If we also have $\nu=\tilde{f}\,d\mu+\lambda$ where $\tilde{f}$ is an extended $\mu$-integrable function, then $$\tilde{f}=f\quad\mu\text{-a.e}$$
Proof.
Case 1
Suppose first that $\mu$ and $\nu$ are both finite, positive measures. On this first step I have no problems
Case 2
Suppose that $\mu$, $\nu$ are both $\sigma$-finite positive measure. We can write $$X=\bigcup_l E_j\quad\text{and}\quad X=\bigcup_k F_k,$$ with $\mu(E_j)<\infty$, $\nu(F_k)<\infty.$ Then $$X=\bigcup_{j,k}(E_j\cap F_k)=\bigcup_l A_l$$ disjointly with $\mu(A_l), \nu(A_l)<\infty.$ Define $$\mu_k(E)=\mu(E\cap A_k)\quad \nu_k(E)=\nu(E\cap A_k),$$ so by case 1. we can write $\nu_k=\rho_k+\lambda_k$ for some unique measures with $\rho_k \ll \mu_k$ and $\lambda_k\perp \mu_k.$ Note that $$\mu_k(A_k^c)=\mu(A_k^c\cap A_k)=0,$$ so $A_k^c$ is a $\mu_k-$null set. Therefore $$f^{'}_k=f_k\chi_{A_k}$$ equals $f_k$ $\mu_k-$ a.e, so we can replace $f_k$ with $f^{'}_k$ without changing $\lambda_k$ or $\rho_k.$ In other words, we can assume that $f_k(x)=0$ $\forall x\notin A_k.$
Since the $A_k$ are disjoint, we can therefore define $$f=\sum_{k=1}^\infty f_k.$$ Since $f\ge 0$, $$d\rho=f\,d\mu$$ defines a positive measure. Also, $$\lambda=\sum_{k=1}^\infty \lambda_k$$ is a positive measure, since each $\lambda_k\ge 0$
Question I'm trying and trying again to show the following but I can't:
- $\lambda, \rho$ are $\sigma$-finite;
- $\nu=\rho+\lambda$;
- $\rho \ll \mu$;
- $\lambda\perp \mu$;
- The uniqueness statements hold.
Could you please give me some suggestions on the basis of what I have already shown?
Best Answer
Let $\mu$ and $\nu$ be two $\sigma$-finite measures. Then we can write $$\bigsqcup_{j=1}^\infty E_j=X=\bigsqcup_{k=1}^\infty F_k,$$ where $\mu(E_j)<\infty,\nu(F_k)<\infty$, for all $k,j$. Let's write $$X=\bigsqcup_{k,j}E_j\cap F_k=\bigsqcup_{l=1} A_l.$$ Then, $\mu(A_l)<\infty,\nu(A_l)<\infty$ for all $l$.
Define two finite measure $\mu_l:\mathcal A\ni E\longmapsto \mu(A_l\cap E)$ and $\nu_l:\mathcal A\ni E\longmapsto \nu(A_l\cap E)$ for each $l\geq 1$. Now, we have $$\nu_l=\lambda_l+\rho_l\text{ for some measures }\lambda_l,\rho_l\text{ with }\lambda_l\perp \mu_l, \rho_l\ll\mu_l\text{ and }\mathrm d\rho_l=f_l\ \mathrm d\mu_l$$$$\text{ for some }\mu_l\text{-integrable real valued non-negative function }f_l.$$
Since $\mu_l(A_l^c)=\nu_l(A_l^c)=0$ we have $\displaystyle\lambda_l(A_l^c)=\mu_l(A_l^c)-\int_{A_l^c}f_l\ \mathrm d\mu_l=0$. We can also assume $f_l=0$ on $A_l^c$.
$\color{red}{\text{Proof of (1):}}$ Now, let $\displaystyle \lambda=\sum_l\lambda_l$ and $\displaystyle f=\sum_lf_l$. Note that $\displaystyle\lambda(A_i)=\sum_l\lambda_l(A_i)=\lambda_i(A_i)<\infty$ as $A_l$ are disjoint, i.e., $\displaystyle A_i\cap \bigsqcup_{l\not= i}A_l=\emptyset$ and $\lambda_l(A_l^c)=0$. So, $\displaystyle X= \bigsqcup_{l=1} A_l$ is a decompostion of $X$ into finite $\lambda$-measure sets.
Define $\mathrm d\rho:=f\ \mathrm d\mu$. Then $$\begin{align}\rho(A_i) &=\int_{A_i} f\ \mathrm d\mu\\ &=\int f1_{A_i}\ \mathrm d\mu\\ & =\int\left(\sum_lf_l 1_{A_i}\right)\ \mathrm d\mu\\ &=\sum_l\int f_l1_{A_i}\ \mathrm d\mu\\ &=\int f_i1_{A_i}\ \mathrm d\mu\end{align}$$ as $f_l(A_l^c)=0$. Now, $\displaystyle\int f_i1_{A_i}\ \mathrm d\mu=\int f_i\ \mathrm d\mu_i<\infty$, see definition of $\mu_i$.
Note that interchange of summation and integral is possible here as all $f_l$ are non-negatives, so apply monotone convergence to partial sums. So, $\displaystyle X= \bigsqcup_{l=1} A_l$ is a decomposition of $X$ into finite $\rho$-measure sets. $\square$
$\color{red}{\text{Proof of (2):}}$ Notice that $\displaystyle\nu=\sum_l\nu_l=\sum_l\lambda_l+\sum_l\rho_l=\lambda+\sum_l\rho_l$. Note that for any $B\in \mathcal A$ we have $\displaystyle B=\bigsqcup_l B\cap A_l$. So, $\displaystyle\rho(B)=\sum_l\rho(B\cap A_l).$ But as in the previous paragraph we have $$\begin{align}\rho(A_i\cap B) &=\int_{A_i\cap B} f\ \mathrm d\mu\\ &=\int f1_{A_i\cap B}\ \mathrm d\mu\\ &=\int\left(\sum_lf_l 1_{A_i\cap B}\right)\ \mathrm d\mu\\ &=\sum_l\int f_l1_{A_i\cap B}\ \mathrm d\mu\\ &=\int f_i1_{A_i\cap B}\ \mathrm d\mu\end{align}$$ as $f_l(A_l^c)=0$. Now, $\displaystyle\int f_i1_{A_i\cap B}\ \mathrm d\mu=\int_B f_i\ \mathrm d\mu_i=\rho_i(B)$.
So, $\displaystyle\rho(A_i\cap B)=\sum_l\rho(B\cap A_l)=\rho_i(B).$ Hence, $\displaystyle\sum_l\rho_l=\rho$. Therefore, $\displaystyle\nu=\lambda+\rho$. $\square$
$\color{red}{\text{Proof of (3):}}$ Note that $\mu(B)=0$ implies $\displaystyle\mu_i(B)=\mu(B\cap A_i)=0$ for all $i$, then $\displaystyle\rho_i(B)=\int_Bf_i\ \mathrm d\mu_i=0$. Since $\displaystyle\rho(B)=\sum_l\rho_l(B)=0$, we are done. $\square$
$\color{red}{\text{Proof of (4):}}$ Since $\mu_l\perp\lambda_l$, write $X=W_l\sqcup U_l$ with $U_l$ is null for $\mu_l$, $W_l$ is null for $\lambda_l$. Let, $\widetilde W_l=W_l\cap A_l,\widetilde U_l=U_l\cap A_l$. Then let $$W=\bigsqcup_l \widetilde W_l, U=\bigsqcup_l \widetilde U_l,$$ see $A_l$ are disjoint. Now, $$W\cap U=\emptyset\text{ and }W\cup U=\bigcup_l\big(\widetilde W_l\cup \widetilde U_l\big)=\bigcup_lA_l=X.$$ Next, $U$ is null for $\mu$ and $W$ is null for $\lambda$ as $\displaystyle\mu(U)=\sum_l\mu_l(U)=\sum_l\sum_j\mu_l(\widetilde U_j)=0$. Similarly for other. $\square$
$\color{red}{\text{Proof of (5):}}$ Now, if $f_l'$ also satisfies the above then, $f_l$ is $\mu_l$ equal to $f_l'$, i.e., $\mu(Z_l\cap A_l)=\mu_l(Z_l)=0$, where $Z_l=\{f_l\not =f_l'\}$.
Note that we can assume, as previously said, $f_l(A_l^c)=0$ and $f_l'(A_l^c)=0$, so that $Z_l\subseteq A_l$, so that $\mu(Z_l)=\mu(Z_l\cap A_l)=0$. Now let $$f'=\sum_lf_l',$$ then $\{f\not=f'\}=\cup_l Z_l$ and so, $\mu\big(\{f\not=f'\}\big)=\sum_l \mu(Z_l)=0$. $\square$