Question regarding Radon-Nikodym derivative... The problems are as follows:

(1) Let $X=[0,1]$ with Lebesuge measure and consider probability measures $\nu,\mu$ given by densities $f,g$ as follows:
  $$\nu(E)=\int_{E} f\;dm\;\;and\;\;\mu(E)=\int_{E}g\;dm$$
  $\forall E \subset [0,1]$ s.t $E$ measurable. Suppose $f(x),g(x)>0$ $\forall x\in [0,1]$. Is $\nu$ absolutely continuos w.rt to $\mu$? If it is, determine the Radon-Nikodym derivative $\frac{d\nu}{d\mu}$. Is $\mu$ absolutely continuos w.r.t to $\nu$?
(2) For a point $x$, define the Dirac measure $\delta_x$ to be
  $$\delta_x=\chi_{A}$$
  where $\chi_{A}$ is the indicator function over set A. For a fixed set $B$, define the Lebesgue measure restricted to $B$ by $m_B(A)=m(A \cap B)$. Let $\mu=\delta_1 + m_{[2,4]}$ and $\nu=\delta_0+m_{(1,2)}$. Show that $\nu$ is singular to $\mu$.

So, here's what I've done:

1) Since $f(x),g(x)>0$ $\forall x \in [0,1]$, it follows that
  $$\nu(E),\mu(E)>0 \iff m(E)>0$$
  and
  $$\nu(E),\mu(E)=0 \iff m(E)=0$$
  $$\implies \nu(E) << \mu(E)\;\;\;,\;\;\; \mu(E) << \nu(E)$$
We wish to find $\frac{d\nu}{d\mu}$. Having $f=\frac{d\nu}{dm}$ and $g=\frac{d\mu}{dm}$, and then dividing $f$ by $g$, we obtain
  $$\frac{\frac{d\nu}{dm}}{\frac{d\mu}{dm}}=\frac{f}{g}$$
  $$\implies \frac{d\nu}{d\mu}=\frac{f(x)}{g(x)}$$
  $\square$
(2) Have
  $$\mu=\delta_1 + m_{[2,4]}\;\;,\;\;\nu=\delta_0+m_{(1,2)}$$
  $$\implies \mu(A)=\delta_1(A)+m_{[2,4]}(A)=\delta_1(A) + m(A\cap [2,4])$$
  $$\nu(A)=\delta_0(A)+m_{(1,2)}(A)=\delta_0(A)+m(A\cap(1,2))$$
  Notice that at each delta, we're dealing with a single point. In particular, the singletons $\{1\}=[1,1]$ and $\{0\}=[0,0]$. Thus,
  $$\mu(A)=\delta_1(A) + m(A\cap [2,4])=\mu(A\cap([1,1]\cup [2,4]))$$
  $$\nu(A)=\delta_0(A)+m(A\cap(1,2))=\nu(A\cap([0,0]\cup (1,2)))$$
  Since $(A\cap([1,1]\cup [2,4])\cap(A\cap([0,0]\cup (1,2)))=\emptyset$, it follows that $\mu$ and $\nu$ are singular to one another.
$\square$

Did I pull some nonsense? Radon and I have not been getting along so well, so I've had trouble properly wrapping my head around these findings. If I've gone down the wrong path, how I would I go about solving these problems?
 A: First is correct: you can also write it shorter. Recall that two measures $\mu$ and $\nu$ are called equivalent if $\mu\ll \nu$ and $\nu\ll\mu$: this is denoted as $\mu\sim \nu$. This is an equivalence relation, and in particular it is transitive. Furthermore, $\mu\sim \nu$ iff $\frac{\mathrm d\nu}{\mathrm d\mu}$ exists and positive  ($\mu$-a.e.) or equivalently iff $\frac{\mathrm d\mu}{\mathrm d\nu}$ exists and positive  ($\nu$-a.e.).


*

*Since $f,g>0$ we have $\mu\sim  m$ and $\nu\sim m$ hence by transitivity $\mu\sim \nu$. The ratio formula is also correct.

*To show mutual singularity, we need to find a set $A$ such that $\mu(A^c) = 0$ and $\nu(A) = 0$: that is, $\mu$ is concetrated on $A$ and $\nu$ is on $A^c$. From the shape of $\mu$ it is clear, that $A = \{1\} \cup [2,4]$ is a good candidate, and you just need to check the latter two conditions to see that it works.
Your solution of 2. has certainly good thoughts, but is a bit confusing and unclear, I'd say. Especially the point where you express $\mu$ through $m$ and then again through $\mu$ itself.
