$d\nu =f\,d\mu$ and $d\mu=g\,d\lambda$ implies $d\nu=fg\,d\lambda$ The full question is

Let $\nu$ be a $\sigma$-finite signed measure, and let $\mu, \lambda$ be $\sigma$-finite 
  positive measures on $(X, \Sigma)$. Show that $\nu \ll \mu$ and $\mu \ll \lambda \implies \nu \ll \lambda.$ Further, show that if $d\nu =f\,d\mu$ and $d\mu=g\,d\lambda$, then $d\nu=fg\,d\lambda$

I found the first part pretty straight forward. However, I'm finding the second part a bit trickier.
I was given the hint of expressing $X$ as a union increasing nested sequence of sets. So $X_1 \subseteq X_2 \subseteq ...$ such that $X = \cup X_n$. Then using simple functions so that $\Psi_k=f*\chi_{X_k}\nearrow f$. Here, $\chi$ is the characteristic function. Then using the monotone convergence theorem for $\int\Psi_k \,d\mu \nearrow \int f\,d\mu$.
I feel like there could be an easier way to show this.
 A: We go to show that $\int\phi\,d\mu=\int\phi g\,d\lambda$ for any
non-negative measurable function $\phi:X\rightarrow[0,\infty]$. Clearly,
if $\phi=1_{A}$ for some $A\in\Sigma$, the equation holds (by the
very definitino of $d\mu=g\,d\lambda$). By linearity on both sides,
the equation holds for all non-negative simple functions $\phi$.
Now let $\phi:X\rightarrow[0,\infty]$ be a measurable function. Then
there exists a sequence of simple functions $(\phi_{n})$ such that
$0\leq\phi_{1}\leq\phi_{2}\leq\ldots\leq\phi$ and $\phi_{n}\rightarrow\phi$
pointwisely. Note that $g\geq0$, so we also have $\phi_{n}g\nearrow\phi g$.
By applying the Monotone Convergence Theorem twice, we have 
\begin{eqnarray*}
\int\phi\,d\mu & = & \lim_{n}\int\phi_{n}\,d\mu\\
 & = & \lim_{n}\int\phi_{n}g\,d\lambda\\
 & = & \int\phi g\,d\lambda.
\end{eqnarray*}
We assert that $\int\phi\,d\mu=\int\phi g\,d\lambda$ for any measurable
function $\phi:X\rightarrow[-\infty,\infty]$ that satisfies the condition:
$\int\phi^{+}\,d\mu<\infty$ or $\int\phi^{-}\,d\mu<\infty$, where
$\phi^{+}=\max(\phi,0)$ and $\phi^{-}=\max(-\phi,0)$.
Proof: Given such $\phi$, we write $\phi=\phi^{+}-\phi^{-}$. Since
at least one of $\int\phi^{+}\,d\mu$, $\int\phi^{-}\,d\mu$ is finite,
we would not encounter $\infty-\infty$. By linearity, we have 
\begin{eqnarray*}
\int\phi\,d\mu & = & \int\phi^{+}\,d\mu-\int\phi^{-}\,d\mu\\
 & = & \int\phi^{+}g\,d\lambda-\int\phi^{-}g\,d\lambda\\
 & = & \int(\phi^{+}-\phi^{-})g\,d\lambda\\
 & = & \int\phi g\,d\lambda.
\end{eqnarray*}
Note that for signed measure $\nu$, it is well-known that $\nu^{+}=f^{+}d\mu$
and $\nu^{-}=f^{-}d\mu$. Moreover, it is required that $\nu^{+}(X)<\infty$
or $\nu^{-}(X)<\infty$.
In particular, for any $A\in\Sigma$, 
\begin{eqnarray*}
\nu(A) & = & \int_{A}f\,d\mu\\
 &  & \int(f1_{A})\,d\mu\\
 & = & \int(f1_{A})g\,d\lambda\\
 & = & \int_{A}fg\,d\lambda,
\end{eqnarray*}
by observing that $\int (f1_A)^+ \,d\mu <\infty$ or $\int (f1_A)^- \,d\mu<\infty$.
That is, $d\nu=fg\,d\lambda$.
