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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 $dnu =f,dmu$ and $dmu=g,dlambda$, then $dnu=fg,dlambda$

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 $Psi_k=f*chi_X_knearrow f$. Here, $chi$ is the characteristic function. Then using the monotone convergence theorem for $intPsi_k ,dmu nearrow int f,dmu$.

I feel like there could be an easier way to show this.

We go to show that $intphi,dmu=intphi g,dlambda$ for any
non-negative measurable function $phi:Xrightarrow[0,infty]$. Clearly,
if $phi=1_A$ for some $AinSigma$, the equation holds (by the
very definitino of $dmu=g,dlambda$). By linearity on both sides,
the equation holds for all non-negative simple functions $phi$.
Now let $phi:Xrightarrow[0,infty]$ be a measurable function. Then
there exists a sequence of simple functions $(phi_n)$ such that
$0leqphi_1leqphi_2leqldotsleqphi$ and $phi_nrightarrowphi$
pointwisely. Note that $ggeq0$, so we also have $phi_ngnearrowphi g$.
By applying the Monotone Convergence Theorem twice, we have
begineqnarray*
intphi,dmu & = & lim_nintphi_n,dmu\
& = & lim_nintphi_ng,dlambda\
& = & intphi g,dlambda.
endeqnarray*

We assert that $intphi,dmu=intphi g,dlambda$ for any measurable
function $phi:Xrightarrow[-infty,infty]$ that satisfies the condition:
$intphi^+,dmu<infty$ or $intphi^-,dmu<infty$, where
$phi^+=max(phi,0)$ and $phi^-=max(-phi,0)$.

Proof: Given such $phi$, we write $phi=phi^+-phi^-$. Since
at least one of $intphi^+,dmu$, $intphi^-,dmu$ is finite,
we would not encounter $infty-infty$. By linearity, we have
begineqnarray*
intphi,dmu & = & intphi^+,dmu-intphi^-,dmu\
& = & intphi^+g,dlambda-intphi^-g,dlambda\
& = & int(phi^+-phi^-)g,dlambda\
& = & intphi g,dlambda.
endeqnarray*

Note that for signed measure $nu$, it is well-known that $nu^+=f^+dmu$
and $nu^-=f^-dmu$. Moreover, it is required that $nu^+(X)<infty$
or $nu^-(X)<infty$.
In particular, for any $AinSigma$,
begineqnarray*
nu(A) & = & int_Af,dmu\
& & int(f1_A),dmu\
& = & int(f1_A)g,dlambda\
& = & int_Afg,dlambda,
endeqnarray*
by observing that $int (f1_A)^+ ,dmu <infty$ or $int (f1_A)^- ,dmu<infty$.

That is, $dnu=fg,dlambda$.