Here is the question:
Consider the space $X = [0,1]$ with Lebesgue measure $\lambda$. Let $\mu = f\lambda$ and $\nu = g\lambda$ with functions $f$ and $g$ nonnegative, be finite measures. Find a condition characterising the absolute continuity $\nu \ll \mu$ and find the Radon Nikodym derivative $\dfrac{\mathrm{d}\nu}{\mathrm{d}\mu}$.
On the first task, $\forall A \in \mathcal L [0,1]$, we have:
$$[\mu(A) = 0 \implies \nu(A) = 0] \iff \\ \big[\int_Afd\lambda = 0 \implies \int_Ag d\lambda = 0 \big] \iff\\ \big[f \chi_A = 0\ (\lambda.a.e)\implies g \chi_A = 0\ (\lambda.a.e) \big].$$
So on whichever sets $f$ is $0$ almost everywhere, g is also $0$ almost everywhere (w.r.t. $\lambda$). Is this a good characterisation? Can you see a better one?
On the second task:
I can guess that we have $\dfrac{\mathrm{d}\nu}{\mathrm{d}\mu} = \dfrac gf$. We have established in lectures the chain rule$$ \frac{\mathrm{d}\nu}{\mathrm{d}\mu} = \frac{\mathrm{d}\nu}{\mathrm{d}\lambda} \frac{\mathrm{d}\lambda}{\mathrm{d}\mu} = g\frac{\mathrm{d}\lambda}{\mathrm{d}\mu}. $$ Now I need an argument that justifies $\dfrac{d\lambda}{d\mu} = f^{-1}\ (\lambda. a.e.)$ "wherever $g \neq 0$". Not sure how to argue in this way.
$\DeclareMathOperator{\supp}{supp}\def\d{\mathrm{d}}$For the first part, the characterization is fine, or it could be expressed as$$ λ(\supp(g) \setminus \supp(f)) = 0. $$
For the second part, because $μ$ and $ν$ are finite measures on $[0, 1]$, then for any $A \in \mathscr{L}([0, 1])$,$$ \int\limits_A g \,\d λ = \int\limits_A \d ν = \int\limits_{[0, 1]} I_A \,\d ν = \int\limits_{[0, 1]} I_A \frac{\d ν}{\d μ} \,\d μ = \int\limits_A \frac{\d ν}{\d μ} \,\d μ = \int\limits_A \frac{\d ν}{\d μ} f \,\d λ. $$ Thus $g = \dfrac{\d ν}{\d μ} f$ holds almost everywhere on $[0, 1]$, which implies $\dfrac{\d ν}{\d μ} = \dfrac{g}{f}$ holds almost everywhere on $\supp(f) \subseteq [0, 1]$.