Another question about proving Lebesgue Decomposition 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?

 A: 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 $\lambda_2$.

