Q:

Question on singular measures and absolute continuity

Below is a question from an old real analysis exam that I need help in solving.
Let $\mu, \nu$, and $\lambda$ be $\sigma$-finite, nonnegative and nontrivial measures on the measure space $(X,\mathcal{M})$ such that for every measurable set $A\in \mathcal{M}$, $$\mu(A) = \nu(A) + \lambda(A).$$ Then I want help in showing the following:  
(i) If $\nu$ and $\lambda$ are mutually singular, then $\exists$ $B\in \mathcal{M}$ such that $\nu(A) = \mu(A\cap B)$.
(ii) If $\nu \ll \lambda$, then $\displaystyle 0\leq \frac{d\nu}{d\mu} \lt 1$ a.e. $[\mu]$ on $X$.  
PS: I have only just learnt signed measures and the Radon-Nikodym Theorem.    

For (i) using @Robert's hints, since $\nu \perp \lambda$, there exists measurable sets $B$ and $B^c$ such that $\nu(B^c) = 0 =\lambda(B).$ Also, for any $A\in \mathcal{M}$, $\nu(A) = \nu(A\cap B)$ and $\lambda(A) = \lambda(A\cap B^c)$. Thus, for any set $A\in \mathcal{M}$, $$
\begin{align*}
\mu(A\cap B) & = \nu(A\cap B) + \lambda(A\cap B)\\
 & = \nu(A) + 0\\
& = \nu(A).
\end{align*}$$

A:

Hint for (i): take $B$ such that $\lambda(B)=0$ and $\nu(B^c) = 0$.
Hint for (2): if $C = \{x: \frac{d\nu}{d\mu}(x) \ge 1$, what can you say about $\nu(C)$, $\mu(C)$ and $\lambda(C)$?

