Folland states that if $\nu$ is a $\sigma-$finite signed measures and $\mu$ is a $\sigma-$finite measure we have:
$$\nu=\nu_c+\nu_s \quad \text{such that $\nu_c<<\mu$ and $\nu_s \perp \mu$.}$$
I understand the proof of this fact but not unicity. If we had $\nu_c+\nu_s=\nu_c'+\nu_s'$ one is tempted to argue $\nu_c-\nu_c'=\nu_s'-\nu_s$ which is both perpendicular and continuous with respect to $\mu$ from which follows unicity.
I cannot understand why $\nu_c+\nu_s=\nu_c'+\nu_s'\Rightarrow \nu_c-\nu_c'=\nu_s'-\nu_s $. How come are we not writing some gibberish like $\infty-\infty$?
By $\sigma$-finiteness there is an increasing sequence of measurable sets $\{A_n\}_n$, whose union is the whole measure space, such that $\nu_{|A_n}$ is finite. If $\nu=\nu_c+\nu_s$ is as in Lebesgue's decomposition theorem we still have ${\nu_c}_{|A_n}<< \mu_{|A_n}$ and ${\nu_s}_{|A_n}\perp \mu_{|A_n}$. You argument shows that ${\nu_c}_{|A_n}={\nu'_c}_{|A_n}$ and ${\nu_s}_{|A_n}={\nu'_s}_{|A_n}$, whence $\nu_c=\lim\limits_n{\nu_c}_{|A_n}=\lim\limits_n{\nu'_c}_{|A_n}=\nu'_c$. Similarly $\nu_s=\nu'_s$.