Exercise 8.O in Bartle's The Elements of Integration I have a doubt about this exercise (8.O) in Bartle's book.
Exercise 8.O

I already answered the Exercise 8.N so I'm able to apply it, but, I just have no idea about how to do this. I'm working on it, but if you guys have any hint to me, would be great.
 A: Let $\lambda,\mu,\nu$ be $\sigma$-finite measures on $(X,\textbf{X})$. Assume first that $\nu\ll\lambda$ and $\lambda\ll\mu$. Take $g=\chi_E\frac{d\mu}{d\lambda}$ in 8.N and let $f=\frac{d\lambda}{d\mu}$ so that$$
\int_Eg\,d\lambda=\int_E gf\,d\mu=\int_E \frac{d\mu}{d\lambda}\frac{d\lambda}{d\mu}\,d\mu.
$$
By the Radon-Nikodym Theorem, we have that$$
\nu(E)=\int_E g\,d\lambda=\int_E \frac{d\mu}{d\lambda}\frac{d\lambda}{d\mu}\,d\mu.
$$
Hence,$$
\frac{d\mu}{d\lambda}\frac{d\lambda}{d\mu}=\frac{d\nu}{d\mu}.
$$
Now, for $j=1,2$, let $\lambda_j\ll\mu$. By the Radon-Nikodym Theorem, we have that$$
(\lambda_1+\lambda_2)(E)=\int_E f\,d\mu.
$$
On the other hand,\begin{align*}
\int_E\left(\frac{d\lambda_1}{d\mu}+ \frac{d\lambda_2}{d\mu}\right)\,d\mu&=\int_E \frac{d\lambda_1}{d\mu}\,d\mu+\int_E\frac{d\lambda_2}{d\mu}\,d\mu=\lambda_1(E)+\lambda_2(E)=(\lambda_1+\lambda_2)(E).
\end{align*}
Now, let $\{\phi_n^\pm\}$ be such that $\phi_n^\pm\nearrow g^\pm$, with $g\in M^+(X,\textbf{X})$. Then,\begin{align*}
\int_Xg\,d\lambda&=\int_X g^+\,d\lambda -\int_Xg^-\,d\lambda=\lim_{n\to\infty}\int_X \phi_n^+\,d\lambda-\lim_{n\to\infty}\int_X\phi_n^-\,d\lambda\\
&=\lim_{n\to\infty}\int_X \phi_n^+f\,d\mu -\lim_{n\to\infty}\int_X\phi_n^- f\,d\mu=\int_X\lim_{n\to\infty}\phi_n^+f\,d\mu-\int_X \lim_{n\to\infty} \phi_n^-f\,d\mu\\
&=\int_X g^+f\,d\mu-\int_X g^-f\,d\mu=\int_X(g^+-g^-)f\,d\mu\\
&=\int_X gf\,d\mu.
\end{align*}
