Can anyone tell me if I am grasping this correctly? Let $(X,\mathcal{F})$ be a measurable space where $X=[0,4\pi]$and then let $\lambda_{1}$ be a signed measure defined by $$ \lambda_{1}(E)=\int_{E} sin(x) \, d\mu(x) $$ for $E \in \mathcal{F}$, where $\mu$ is the Lebesgue measure. Now define another signed measure $\lambda_{2}$ defined by $$ \lambda_{2}(E)=\int_{E}cos(x) \, d\mu(x) $$ for $E \in \mathcal{F}$. Now for $E=[0,2\pi]$ we have $$ \lambda_{1}([0,2\pi])=\int_{[0,2\pi]} sin(x) \, d\mu(x)=\int^{2\pi}_{0} sin(x) \, dx=0 $$ and then for $E=(2\pi,4\pi]$ we have $$ \lambda_{2}((2\pi,4\pi])=\int_{(2\pi,4\pi]}cos(x) \, d\mu(x)=\int^{4\pi}_{2\pi} cos(x) \, dx=0$$ and then we have that $[0,2\pi] \cap (2\pi,4\pi]=\emptyset$ and $[0,2\pi] \cup (2\pi,4\pi]=X$ which seems to satisfy all of the criteria for mutual singularity (i.e. $\lambda_{1} \perp \lambda_{2}$) but the fact that I could easily reverse this suggest I am either missing something or that I have a trivial/uninteresting example. Can anyone point me in the direction of a more interesting example if this is the case? Also if this is an elementary/stupid question, I apologize in advance.
Major Edit: Now that I've read the entire section, I think that I have a much better understanding of this concept. Here is a new example that reflects my current understanding. Let $(X,\mathcal{F})$ be a measurable space as above and let $(A,B)$ be a Hahn decomposition with respect to a signed measure $$ \lambda(E)=\int_{E} sin(x) \, d\mu(x) $$ then one Hahn decomposition of $X$ is given by $A=[0,\pi)\cup[2\pi,3\pi)$ and $B=[\pi,2\pi) \cup [3\pi,4\pi]$ since Hahn decompositions are essentially unique. Now let $\mu(E)=\lambda(E \cap A)$ and $\nu(E)=-\lambda(E \cap B)$ for $E \in \mathcal{F}$. Now, let's say that $E=(0,3\pi)$ and clearly $(0,3\pi) \in \mathcal{F}$ . Then $\mu(E)=\lambda((0,3\pi)\cap(([0,\pi)\cup[2\pi,3\pi)))$ which is given by $$ \mu(E)=\int^{\pi}_{0} sin(x) \, dx +\int^{3\pi}_{2\pi} sin(x) \, dx=4 $$ and $\nu(E)=-\lambda((0,3\pi)\cap([\pi,2\pi)\cup[3\pi,4\pi]))$ which is given by $$ \nu(E)=-\int^{2\pi}_{\pi} sin(x) \, dx =2 $$ and since $\lambda(E)$ is given by $$ \lambda(E)=\int^{3\pi}_{0} sin(x) \, dx=2 $$ we have that $\lambda=\mu-\nu$ and since $$ \mu(B)=\lambda(B \cap A)=0 $$ since $A \cap B =\emptyset$ and then for that same reason, we have that $$ \nu(A)=-\lambda(A \cap B)=0 $$ which implies that $\mu \perp \nu$ and that $(\mu,\nu)$ is a Jordan decomposition for $(X,\mathcal{F},\lambda)$. Is this example correct? Also, is my understanding correct that the whole point of the Jordan decomposition is just to get two positive measures who add to give $\lvert \lambda \rvert=\mu + \nu$ and $\lambda=\mu-\nu$ in the same way we write $f^{+}=\max\{f(x),0\}$ and $f^{-}=\max\{-f(x),0\}$ so that $f=f^{+}-f^{-}$ and $\lvert f \rvert=f^{+}+f^{-}$? I'm just now starting to get into the section on variation of measures but based on my skimming of that section, this seems to be the case. I apologize if I should have made this into a new question but I do not want to clog up the system.
You need $\lambda_1$ equal to zero on all measurable subsets of $[0,2\pi]$, not only on $[0,2\pi]$ (same with $\lambda_2$ on $]2\pi, 4\pi]$)