3

Note: This is my original question. I have been kindly helped to turn this into a correct proof, which I have posted as an answer so this question won't show up as "unanswered".
As an exercise, I am trying to provide a rigorous proof of uniqueness in the Lebesgue Decomposition Thm, assuming we already have existence. I am following the outline provided here. Below is what I have so far
Step 1: Assume $\lambda$ is a finite measure ($\mu$ only needs to be $\sigma$-finite).

  • Let $$\lambda=\lambda_1+\lambda_2 = \lambda_3 + \lambda_4, \text{where } \lambda_1, \lambda_3 \perp \mu \text{ and } \lambda_2,\lambda_4 \ll \mu \tag1$$
  • Let $\alpha:=\lambda_3 -\lambda_1 =\lambda_2-\lambda_4$
  • Extend defns of singular & abs cont to signed measures: (I just do the most intuitive thing here) $\alpha \perp \mu $ means $\exists$ a partition A,B of X s.t. $\alpha(A)= \mu(B)=0$. $\alpha \ll \mu$ means $\mu(E)$=0 implies $\alpha(E)$=0.
  • Show $\alpha \perp \mu$ and $\alpha \ll \mu$. This is just checking defns.
  • Conclude $\alpha$=0. I'm stuck here. By the previous pt, $\exists$ a partition A,B of X s.t. $\alpha(A)=\mu(B)$=0. Further, $\mu(B)$=0 implies $\alpha(B)=0$. So $\alpha(X)=\alpha(A)+\alpha(B)$=0. But we may have a partition $X_1,X_2$ of X s.t. 0 $< \alpha(X_1)=-\alpha(X_2)$.

Step 2: (General case) $\lambda$ is $\sigma$-finite.

  • Let $X_1 \subseteq X_2 \subseteq$..., $X=\cup_{n=1}^{\infty} X_n$, each $\lambda(X_n)< \infty$.
  • $\forall n$, $E \in \textbf{X}$, put $\lambda_n(E):=\lambda(E \cap X_n)$, which is a finite measure, so $\exists$ a unique Lebesgue decomp $\lambda_n=\lambda_{1n}+\lambda_{2n}$
  • Assume (1) again. Now I don't know how to show $\lambda_1=\lambda_3$. I’m trying to use the previous bullet pt. I think I can show $\lambda=(\lim_{n \to \infty} \lambda_{1n})+ (\lim_{n \to \infty} \lambda_{2n})$ is a Lebesgue Decomp, but how do I know there aren’t others?
Community
  • 1
Jason
  • 2,065
  • 1
    Say $E$ is $\nu$-null if $\nu(F)=0$ for every $F\subset E$. For signed measures this is not the same as just $\nu(E)=0$. Go through your proof (including the definitions), and change "$\nu(E)=0$" to "$E$ is $\nu$-null" everywhere and see what happens. (One of the new definitions is equivalent to the old one, one is not.) – David C. Ullrich Jun 14 '16 at 22:45
  • Previous comment is something you need to change to make it all work. This one doesn't really matter. But in Step 2 it might work out nicer if you took the $X_n$ to be disjoint instead of nested. – David C. Ullrich Jun 14 '16 at 22:47
  • @DavidC.Ullrich Thx for your helpful comments (both here and on my previous question). I see how to do the $\lambda$ is finite case now. – Jason Jun 14 '16 at 23:52
  • But you still have the wrong definitions in the new version! Saying $\alpha\perp\mu$ simply does not mean what you say it does. You're trying to prove something false. – David C. Ullrich Jun 15 '16 at 00:04
  • Hmm, a possible reason for the confusion: I say that you should say that $E$ is a $\nu$-null set if $\nu(F)=0$ for every $F\subset E$. This is equivalent to saying $|\nu|(E)=0$. But the notation $|\nu|$ can be confusing; $|\nu|(E)$ is not $|\nu(E)|$. (So what is $|\nu(E)|$? Look it up somewhere; it's the "total variation" measure of $\nu$. I didn't mention it because we don't really need it here; I'm mentioning it now because maybe it happened that you've seen these definitions phrased in terms of $|\nu|$; if you have the reasonable but wrong idea that $|\nu|(E)=|\nu(E)|$ this won't work.) – David C. Ullrich Jun 15 '16 at 00:09
  • @DavidC.Ullrich. You're right, I originally had the correction at the bottom, but I think correcting the main part makes it easier on other people viewing the question. – Jason Jun 15 '16 at 00:09
  • I think you're right about that - fooled me in any case... – David C. Ullrich Jun 15 '16 at 00:12
  • Do you happen to know Step 2? I'm sorry to say I'm not seeing the uniqueness :( – Jason Jun 15 '16 at 00:14
  • Well, so far there's no evidence that you've tried doing as I suggested, taking the $X_n$ to be disjoint. Try it. – David C. Ullrich Jun 15 '16 at 00:15
  • That's what I'm trying. So in 1st bullet pt assume the $X_n$ are disjoint. 2nd bullet pt unchanged. 3rd bullet pt all I can show is $\lambda= \sum_{n=1}^\infty \lambda_{1n}+ \sum_{n=1}^\infty \lambda_{2n}$. This is A Lebesgue Decomp, but how do I know it's THE Lebesgue Decomp? For instance, there could be some other decomp constructed from a totally different method. – Jason Jun 15 '16 at 00:23
  • 1
    Say you have another decomposition. The "restriction" of that decomposition to $X_n$ is a decomposition of the "restriction" of $\lambda$ to $X_n$. So the restriction of the decomposition to $X_n$ is uniquely determined. Take $\sum_n$. – David C. Ullrich Jun 15 '16 at 00:35
  • I think I finally got it. Thanks for all your help! – Jason Jun 15 '16 at 01:09

1 Answers1

1

I want to acknowledge David C. Ullrich's role in helping me reach this answer.
Step 1: Assume $\lambda$ is a finite measure ($\mu$ only needs to be $\sigma$-finite).

  • Let $$\lambda=\lambda_1+\lambda_2 = \lambda_3 + \lambda_4, \text{where } \lambda_1, \lambda_3 \perp \mu \text{ and } \lambda_2,\lambda_4 \ll \mu \tag1$$
  • Let $\alpha:=\lambda_3 -\lambda_1 =\lambda_2-\lambda_4$
  • Extend defns of singular & abs cont to signed measures: $\alpha \perp \mu$ means $\exists$ partition A,B of X s.t. $\mu(B)$=0 and A is $\alpha$-null ($\alpha(E \cap A)$=0 $\forall E \in \textbf{X}$). $\alpha \ll \mu$ means $\mu(E)$=0 implies $\alpha(E)$=0.
  • Show $\alpha \perp \mu$ and $\alpha \ll \mu$. This is just checking defns.
  • Conclude $\alpha$=0.

Step 2: (General case) $\lambda$ is $\sigma$-finite.

  • Let $X_n$ be disjoint, $X=\cup_{n=1}^{\infty} X_n$, each $\lambda(X_n)< \infty$.
  • $\forall n$, $E \in \textbf{X}$, put $\lambda_n(E):=\lambda(E \cap X_n)$, which is a finite measure, so $\exists$ a unique Lebesgue decomp $\lambda_n=\lambda_{1n}+\lambda_{2n}$
  • Let $\lambda=\lambda_1+\lambda_2$ be any Lebesgue decomp. Then the measures $E \mapsto \lambda_1 (E \cap X_n)$, $E \mapsto \lambda_2 (E \cap X_n)$ are $\perp, \ll \mu$, resp. So they form a Lebesgue decomp of $\lambda_n$, and by uniqueness, $\lambda_1 (E \cap X_n) = \lambda_{1n} (E)$ and same using $\lambda_2$ instead.
  • $\forall E \in \textbf{X}$: $$\lambda_1 (E) = \sum_{n=1}^\infty \lambda_1 (E \cap X_n) = \sum_{n=1}^\infty \lambda_{1n} (E)$$ where 1st "=" is countable additivity. This shows $\lambda_1$ is uniquely determined. Again, same idea for $\lambda_2$.
Jason
  • 2,065