Is the Lebesgue decomposition only meaningful for measures which are supported on different sets?

Question

Lately I've been trying to understand why we concern ourselves with the Lebesgue decomposition. The only satisfactory answer I can think of is that it allows us to work more easily with "weird" measures. For example, say we have a measurable space $(X,\mathcal{F})$, a positive measure $\nu$, a signed measure $\lambda$ and two signed measures $\lambda_{a}$ and $\lambda_{s}$ such that $\lambda=\lambda_{a}+\lambda_{s}$, $\nu \perp \lambda_{s}$ and $\lambda_{a} \ll \nu$. Say $\mu$ is the Lesbesgue measure and $X=\mathbb{R}$ and for $E \in \mathcal{F}$ we have $$ \nu(E)=\int_{E}g \, d\mu $$ and $$ \lambda(E)=\int_{E} f \, d\mu $$ where $$ g(x)= \begin{cases} 1 & \text{ if }x \in (-\infty,0] \\ 0 & \text{ otherwise} \end{cases} $$ and $$ f(x)=\begin{cases} x^{2} & \text{ if }x \in (-\infty,1] \\ 0 & \text{ if } x \in (1,\infty) \end{cases}$$ which can be decomposed into $f_{1}+f_{2}$ given by $$ f_{1}(x)=\begin{cases} x^{2} & \text{ if }x \in (-\infty,0] \\ 0 & \text{ otherwise} \end{cases}$$ and $$ f_{2}(x)=\begin{cases} x^2 & \text{ if }x \in (0,1] \\ 0 & \text{ otherwise} \end{cases} $$ then define $$ \lambda_{a}(E)=\int_{E} f_{2}\, d\mu $$ and $$ \lambda_{s}(E)=\int_{E}f_{1} \, d\mu $$ clearly we have that $\nu \perp \lambda_{s}$ and $\lambda=\lambda_{a}+\lambda_{s}$ and also that $\lambda_{a} \ll \nu$. So is this only useful when we consider decompositions w.r.t. measures with "weird" supports? For example consider the same example but where $$ \nu(E)=\int_{X} \mathbb{1}_{E} \, d\mu $$ or simply that $\nu(E)=\mu(E)$. Then absolute continuity for $f$ becomes easy since any set $E$ such that $\nu(E)=0$ we will have $$ \lambda(E)=\int_{E} f \, d\mu=0 $$ and then we have that $\lambda=\lambda_{a}+0$ since the Lebesgue measure is supported over all of the real numbers. Would this also be true for any measure supported on $\mathbb{R}$? I'm just having a hard time seeing where this fits in to what I've been studying. I'm somewhat new to measure theory so maybe I just lack the experience but I also worry that I am missing a key piece of insight.

Answer 1

For your first question:

If two measures share the same support then the singular component of the Lebesgue decomposition of one measure w.r.t. to another measure is always zero?

No, this is not necessarily true. Let $m$ denote Lebesgue measure on $[0,1]$ and $\mu$ be a Dirac measure on $[0,1]$ for the point $x = 1/2$. That is $$ \mu(A) = \left\{ \begin{array}{cl} 1 & \text{ if } 1/2 \in A \\ 0 & \text{ if } 1/2 \notin A \end{array} \right.,\ A \in [0,1] \text{ Borel}. $$ Define $\nu := m + \mu$ on the Borel sets of $[0,1]$. What is the Lebesgue decomposition of $\nu$ with respect to $m$?

For for your second question:

If $\nu$ and $\mu$ are measures and we want to find the Radon Nikodym derivative of $\nu$ w.r.t. $\mu$ then the support of $\nu$ can be strictly smaller than that of $\mu$ and still have absolute continuity, right?

Yes. Suppose $\mu$ denotes Lebesgue measure on $\mathbb{R}$ and let $\nu(A) := \mu(A\cap [0,1])$ for all Borel $A\subseteq \mathbb{R}$. Then if $\mu(A) = 0$, $\nu(A) \leq \mu(A) = 0$. Hence $\nu \ll \mu$.