Problem 10J asks many things of us. Other Math Stack Exchange posts such as this one, focus only in one of the aspects of the problem. I will prove all of them for the sake of completeness.
Part I: $E\in X\times Y$ is measurable if and only if $E_n$ is measurable in $Y$.
Proof:
$(\Rightarrow )$ This holds for general spaces and Bartles proves it in Lemma 10.6. However, let me prove it again, because Bartle omits a couple of steps.
Let $\mathcal{R}$ denote the set of all sets in the cartesian product, such that $E_n$ is measurable. Clearly, if $\dot{\cup}_{i=1}^n A_{i}\times B_i$ is a general element in $Z_o$ which is finite union of measurable rectangles:
$$(\dot{\cup}_{i=1}^n A_{i}\times B_i)_x=\cup_{i| x\in A_i}B_i$$
Finite union of measurable sets is measurable in $Y$. But this means that $Z_o\subset \mathcal{R}$. Furthermore $ \mathcal{R}$ is a sigma algebra because if $R_i\in \mathcal{R}$, we have that $(\cup_{i\in \mathbb{N}} R_i)_x=\cup_{i\in \mathbb{N}} (R_i)_x$ which is measurable, so $\cup_{i\in \mathbb{N}} R_i\in \mathcal{R}$. Similarly, if $R_i \in \mathcal{R}$, we have $(R_i^C)_x=({R_i}_x)^C$ to be measurable hence $ R_i \in \mathcal{R}$. Therefore, $\mathcal{R}$ is a sigma algebra containing $Z_o$ and $\textbf{Z}=\sigma(Z_o)\subset \mathcal{R}$
$(\Leftarrow)$ It holds that $E=\cup_{n=1}^\infty \{n\}\times E_n$. Each $\{n\}\times E_n$ is a measurable rectangle by our hypothesis and therefore, countable union of them is certainly measurable.
Part II: There is a unique product measure $\pi$ such that it satisfies:
$$ \pi(E)=\sum_{n=1}^\infty \mu(E_n)$$
Proof: Theorem 10.4 from Bartle establishes (as a consequence of the monotone convergence theorem and Caratheodory), that there is a measure $\pi$ such that $\pi(A\times B)=\mu(A)\nu(B)$, if $A\times B\in Z_o$. However, for any general $E\in \textbf{Z}$, because of part I, we have $E_n$ to be measurable. Hence:
$$\pi(E)=\pi(\dot{\cup}_{n=1}^\infty \{n\}\times E_n )=\sum_{n=1}^\infty \pi(\{n\}\times E_n)=\sum_{n=1}^\infty \nu (E_n)$$
But this establishes the uniqueness part and also the formula to compute $\pi$ in any given set.
Part III: A function $f:X\times Y\rightarrow \mathbb{R}$ is measurable if and only if we have $f_n(y)=f(n,y)$ measurable as a function from $Y$ to $\mathbb{R}$.
Proof: Let $O$ be an open set in $\mathbb{R}$.
$(\Rightarrow)$ $f_n^{-1}(O)=\{y|f(n,y)\in O \}=\{y| (n,y)\in f^{-1}(O)\}=(f^{-1}(O))_n$. But $f^{-1}(O)$ is measurable in $\textbf{Z}$ and therefore, from part I, we infer that $(f^{-1}(O))_n$ is measurable in $\textbf{Y}$.
$(\Leftarrow)$ $f^{-1}(O)=\cup_{n=1}^\infty \{n\}\times f^{-1}_n(O)$. This is clearly in the sigma algebra, because each $\{n\}\times f^{-1}_n(O)\in \textbf{Z}$.
Part IV: $f$ is integrable with respect to $\pi$ if and only if the series:
$$\sum_{n=1}^\infty \int_Y |f_n| d\nu<\infty$$
In this case, we have:
$$\int_Z f d\pi = \sum_{n=1}^\infty \int_Y f d\nu =\int_Y \sum_{n=1}^\infty f d\nu$$
Proof: We note that $f$ is integrable if and only if $|f|$ has a finite integral. But we have simple functions $\varphi_n$ which converge to $|f|$ almost everywhere from which it follows that:
$$\int|f|d\mu=\lim_m \int \varphi_m d\mu=\lim_m \int \sum_{j=1}^{J_m} x_j \chi_{E_{jm}} d\mu=\lim_m \left(\sum_{j=1}^{J_m}x_j\pi(E_{jm})\right)=\lim_m \left(\sum_{j=1}^{J_m} x_j \sum_{n=1}^\infty \nu((E_{jm})_n)\right)=\lim_m \left(\sum_{j=1}^{J_m} x_j \sum_{n=1}^\infty \int\chi_{(E_{jm})_n} d\nu\right)=\lim_m \left(\sum_{j=1}^{J_m} x_j \sum_{n=1}^\infty \int\chi_{E_{jm}}(n,\cdot) d\nu\right)=\lim_m\left( \sum_{n=1}^\infty \int\sum_{j=1}^{J_m} x_j\chi_{E_{jm}}(n,\cdot) d\nu\right)$$
However, we have that $\sum_{j=1}^{J_m} x_j\chi_{E_{jm}}(n,\cdot)=\varphi_m(n,\cdot)\xrightarrow[]{m} |f(n,\cdot)|=|f_n(\cdot)|$ pointwise and monotonically. Therefore,by the monotone converge theorem, we have:
$$\int |f| d\pi =\lim_m\left( \sum_{n=1}^\infty \int \varphi(n,\cdot) d\nu\right)= \sup_m \sup_l \sum_{n=1}^l \int\sum_{j=1}^{J_m} x_j\chi_{E_{jm}}(n,\cdot) d\nu =\sup_l \sup_m \sum_{n=1}^l \int\sum_{j=1}^{J_m} x_j\chi_{E_{jm}}(n,\cdot) d\nu =\sup_l \sum_{n=1}^l\int |f_n| d\nu=\sum_{l=1}^\infty \int |f_n| d\nu$$
The reason why we can exchange the supremums follows from this other MSE post. Therefore, we have the condition we were looking for:
$$\int |f|d\pi<\infty \Leftrightarrow \sum_{i=1}^\infty \int |f_n|d\nu<\infty\quad \text{because} \quad \int |f|d\pi= \sum_{i=1}^\infty \int |f_n|d\nu $$
In this case, me may conclude using the same approximation by simple $\varphi_n^+\rightarrow f^+$ and also $\varphi_n^- \rightarrow f^-$ (and subsequently using the monotone convergence theorem to commute the series and the integration) that:
$$\int f^+d\pi=\sum_{i=1}^\infty \int f_n^+ d\nu= \int \sum_{i=1}^\infty f_n^+d\nu<\infty $$
$$ \int f^-d\pi=\sum_{i=1}^\infty \int f_n^- d\nu= \int \sum_{i=1}^\infty f_n^- d\nu<\infty$$
Subtracting these two equalities, we have the desired conditions.
