Proving uniqueness in Lebesgue decomposition

Question

I'm reading a book(Measure, Integral and Probability by Capinski), where in the proof of the Lebesgue decomposition theorem, it leaves to the reader to prove uniqueness. As a hint, the authors state that we should use the following proposition:

«Let $\mu,\nu, \lambda_1, \lambda_2$ measures on sigma algebra $\mathcal{F}$. Then we have:

i) If $\lambda_1 \perp \mu$ and $\lambda_2 \perp \mu$, then $\lambda_1+\lambda_2 \perp \lambda_2$.

ii) If $ \lambda_1 \ll \mu$, and $\lambda_2 \perp \mu$, then $\lambda_1 \perp \lambda_2$.

iii) If $ \nu \ll \mu$, and $\nu \perp \mu$, then $\nu=0$.»

I've tried using point iii) to the subtraction of the absolutely continuous part of the two representations of the same measure, but subtraction of two measures may not necessarily be a measure...

Any help would be appreciated.

Answer 1

Let $\lambda = \lambda_a + \lambda_s$ be the Lebesgue decomposition of $\lambda$ with respect to the positive finite measure $\mu$, i.e. $\lambda_a \ll \mu$ and $\lambda_s \bot \mu$.

To show uniqueness, take another pair $(\lambda_a', \lambda_s')$ satisfying those properties. Then $$\lambda = \lambda_a + \lambda_s = \lambda_a' + \lambda_s',$$ so \begin{equation}\tag{1}\lambda_a' - \lambda_a = \lambda_s - \lambda_s'.\end{equation}

First of all, $\lambda_a \ll \mu$ and $\lambda_a' \ll \mu$ imply $\lambda_a' - \lambda_a \ll \mu$.

Next, $\lambda_s' \;\bot\; \mu$ and $\lambda_s \;\bot\; \mu$ imply $\lambda_s - \lambda_s' \;\bot\; \mu$.

So, $\mu \; \bot \; \lambda_s - \lambda_s' = \lambda_a' - \lambda_a \ll \mu$ implies that both sides of (1) are equal to 0 by your property (iii).

I'm not sure why the author is giving you that hint, the proofs in the textbooks I've seen all used the following property

If $\lambda_1 \ll \mu$ and $\lambda_2 \ll \mu$, then $\lambda_1 + \lambda_2 \ll \mu$.

If you want to extend this to $\sigma$-finite measure, just take some disjoint sequence $E_n$ in your $\sigma$-algebra such that $X = \bigcup_n E_n$, and consider the restriction $\lambda_n$ of $\lambda$ for each $n \in \mathbb{N}$.

Answer 2

Consider any two such Lebesgue decompositions:

\begin{align*} \nu = \nu_r + \nu_s = \nu_r' + \nu_s' \\ \end{align*}

(with $\nu_r \ll \mu, \nu_r' \ll \mu$ and $\nu_s \bot \mu, \nu_s' \bot \mu$)

Define the signed measure $\alpha$ with the Jordan decomposition $\alpha = \alpha_+ - \alpha_-$ such that $\alpha_+,\alpha_-$ are unsigned measures.

\begin{align*} \alpha = \nu_r - \nu_r' = \nu_s' - \nu_s \\ \end{align*}

Proof that $\alpha_+ \ll \mu$ and $\alpha_- \ll \mu$

Given any Hahn decomposition for signed measure $\alpha$, $\Omega = A \cup B$, defines $A,B$ as positive/negative sets with respect to $\alpha$, then the Jordan decomposition defines for any $E \subset \Omega$ the non-negative unsigned measures $\alpha_+(E) = \alpha(E \cap A), \alpha_-(E) = -\alpha(E \cap B)$ with $\alpha(E) = \alpha_+(E) - \alpha_-(E)$. If $E \subset \Omega$ is such that for any $E' \subset E \Rightarrow \alpha(E') = 0$, then clearly $\alpha_+(E) = \alpha(E \cap A) = 0$ and $\alpha_-(E) = -\alpha(E \cap B) = 0$.

If $E$ is any set such that $\mu(E) = 0$, then clearly for any subset $E' \subset E$, $\mu(E') = 0$, then by $\nu_r \ll \mu$ and $\nu_r' \ll \mu$ we have $\nu_r(E) = \nu_r(E') = \nu_r'(E) = \nu_r'(E') = 0$ and then $\alpha(E') = \nu_r(E') - \nu_r'(E') = 0$ and $\alpha(E) = \nu_r(E) - \nu_r'(E) = 0$ then by our earlier claim $\alpha_+(E) = \alpha_-(E) = 0$. Since this is for any $E$ such that $\mu(E) = 0$ we conclude that $\alpha_+ \ll \mu$ and $\alpha_- \ll \mu$.

Main Proof Continued

Since $\nu_s \bot \mu$ and $\nu_s' \bot \mu$ then, we can choose $A,B$ such that $\nu_s(A) = \nu_s'(B) = \mu(A^c) = \mu(B^c) = 0$. Then $\mu((A \cap B)^c) = \mu(A^c \cup B^c) \le \mu(A^c) + \mu(B^c) = 0$. And $\nu_s' - \nu_s = \alpha(A \cap B) = 0$. Since for any measurable subset $E \subset A \cap B$, then $\alpha(E) = 0$, then $\alpha_+(A \cap B) = \alpha_-(A \cap B) = 0$ and therefore we have both $\alpha_+ \bot \mu$ and $\alpha_- \bot \mu$.

Since $\alpha_+ \ll \mu$ and $\alpha_+ \bot \mu$, then $\alpha_+ = 0$.

Since $\alpha_- \ll \mu$ and $\alpha_- \bot \mu$, then $\alpha_- = 0$.

Therefore $\alpha = \alpha_+ - \alpha_- = 0$ so $\nu_r = \nu_r'$ and $\nu_s = \nu_s'$. $\square$