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Lebesgue's Decomposition Thm states: if $\lambda,\mu$ are $\sigma$-finite measures on a measurable space $(X,\textbf{X})$, then $\exists$ unique measures $\lambda_1,\lambda_2$ on $(X,\textbf{X})$ s.t.

  • $\lambda=\lambda_1+\lambda_2$
  • $\lambda_1 \perp \mu$ (mutually orthogonal: $\exists A,B \in \textbf{X}$ that partitions X, and $\lambda_1(A)= \mu(B)=0$
  • $\lambda_2 << \mu$ (abs cont: $E \in \textbf{X}, \mu(E)=0$ implies $ \lambda_2(E)=0$)

I have read Bartle's pf of this fact in his "Elements of Integration". I'm comfortable with the existence part, but my question is about the uniqueness part.
Bartle says it follows from the fact that if $\alpha$ is a measure s.t. $ \alpha \perp \mu$ and $\alpha << \mu$, then $\alpha$=0.
This post gives an answer that deals with signed measures, starting with finite measures and then extending to the $\sigma$-finite case, but I suspect there is an easier way since I don't feel Bartle would have "buried that much work" inside one sentence, given the nature of his previous proofs. So I guess my question is if there's a way to avoid signed measures.
What I've tried so far is essentially in the link above. Let (since we already have existence)

  • $\lambda=\lambda_1+\lambda_2 = \lambda_3 + \lambda_4$
  • $\lambda_1, \lambda_3 \perp \mu$
  • $\lambda_2,\lambda_4 << \mu$

I'm tempted to use $\alpha= \lambda_3 - \lambda_1$, but we don't know that this is a measure.

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Jason
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  • That's the right way to go. You can prove it's a measure by definition. – Ningxin Jun 12 '16 at 23:04
  • @QiyuWen No, you're missing a point. The difference of two (positive) measures need not be a signed measure, in fact it can be undefined. Because $\infty-\infty$ is undefined... – David C. Ullrich Jun 12 '16 at 23:17
  • I doubt you're going to find a proof that's much simpler. – David C. Ullrich Jun 12 '16 at 23:19
  • @DavidC.Ullrich is it what you meant : Jason's easier way is not so easier because for making sense to $\alpha = \lambda_3 - \lambda_1$ we'll need the same idea of restriction to a sub $\sigma$-algebra where $\lambda_1,\lambda_3$ are finite measures ? – reuns Jun 12 '16 at 23:45
  • @user1952009 I didn't mean that exactly - he didn't say what his easier way was. If we're going to derive the result by considering $\alpha=\lambda_3-\lambda_1$ then yes, it seems we need to start with finite measures. It seems to me he understands this. There could be some totally different approach that would be simpler. I doubt it (because the standard argument really isn't that complicated, for one thing). – David C. Ullrich Jun 12 '16 at 23:50
  • @DavidC.Ullrich Thanks for completion. In the setting of Lebesgue's decomposition I automatically assumed that $\lambda$ was a complex measure. – Ningxin Jun 13 '16 at 00:05
  • The lemma that $\alpha \bot \mu$ and $\alpha \ll \mu$ imply that $\alpha = 0$, does not hold when $\alpha$ is a signed measure. – clay Jun 27 '22 at 21:15
  • I found a proof that does not deal with signed measures (although the document still included the difference of measures). – ムータンーオ Aug 17 '22 at 15:00

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