Fix a measure space $(S,\Sigma,\mu)$ and let $(E_n, n\geq 1)$ be a sequence of disjoint measurable sets so that $\mu(E_n)<\infty$ for all $n$. Now fix another measure $$\nu(A)=\sum_{n=1}^{\infty} 2^{-n} \frac{\mu(A\cap E_n)}{\mu(E_n)+1}$$ for $A\in\Sigma$.
Now if we wish to find two integrable functions $f,g$ that satisfy $\mu(A)=\int g\mathrm{d}\nu$ and $\nu(A)=\int f\mathrm{d}\mu$, then some tedious algebraic reasoning shows that $$ f=\sum_{n=1}^{\infty}2^{-n}\frac{1_{E_n}}{\mu(E_n)+1}\quad \text{and}\quad g=\sum_{n=1}^{\infty}2^n(\mu(E_n)+1)1_{E_n} $$ do the trick. For completeness, if we let $g=\sum_{n=1}^{\infty}\beta_n1_{E_n}$, then $1_Ag=\sum_{n=1}^{\infty}\beta_n 1_{A\cap E_n}$ and hence \begin{equation} \begin{split} \mu(A)&=\int 1_Ag\mathrm{d}\nu\\ &=\sum_{n=1}^{\infty}\beta_n\nu(A\cap E_n)\\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\beta_n2^{-m} \frac{\mu(A \cap E_n\cap E_m)}{\mu(E_m) + 1}\\ &=\sum_{n=1}^{\infty}\beta_n2^{-n} \frac{\mu(A \cap E_n)}{\mu(E_n) + 1} \end{split} \end{equation} hence, if $\beta_n=2^n\mu(E_n)+2^n$, then \begin{equation} \begin{split} \sum_{n=1}^{\infty}\beta_n2^{-n}\frac{\mu(A \cap E_n)}{\mu(E_n) + 1}&=\sum_{n=1}^{\infty}\mu(A\cap E_n)\\ &=\mu(A) \end{split} \end{equation} as desired. A similar argument yields the result for $f$.
Now our argument is complete if we can argue that $f$ and $g$ are indeed integrable. For $f$, observe that $\nu(A)$ is finite for every $A\in\Sigma$ and $\nu(A)=\int_Af\mathrm{d}\mu$ making $f$ integrable.
Now I am stuck at showing that $g$ is also integrable. My gut feeling tells me that $g$ is not integrable as we can not say anything about $\mu(A)$ being finite for every $A\in\Sigma$. Yet, the notes that were provided to us in class do insist on finding integrable functions $f,g$ satisfying the above and so I am wondering if there is perhaps a mistake in my work, or that the exercise contains an error?
For some additional notes: we say that a (positive) function $f$ is integrable with respect to a measure $\mu$ if $f$ is ($\Sigma$)-measurable and if $\mu(f)=\int f\mathrm{d}\mu<\infty$.