Here is the necessary context behind remark 2.8.
Let us prove another important property of the Lebesgue integral: it is a $\sigma$-additive function on $\mathcal{F}.$ Namely, we have the following. Theorem 2.7. Let $B=\bigcup\limits_{i=1}^{\infty} B_i$, where $B_i \in \mathcal{F}, i \ge 1,$ be such that $B_i \cap B_j = \emptyset$ for $i \ne j.$ Let $\xi$ be a random variable such that$\int_B \xi dP$ is finite. Then $\int_B \xi dP =\sum\limits_{i=1}^{\infty} \int_{B_i} \xi dP.$ (2.3)
Proof. Assume, first, that $\xi$ is a simple function given by (2.1). Then $$\int_B \xi dP = \sum\limits_{j=1}^{\infty} x_j P(A_j \cap B) \sum\limits_{i=1}^{\infty}\sum\limits_{j=1}^{\infty} x_j P(A_j \cap B_i) =\sum\limits_{i=1}^{\infty} \int_{B_i} \xi dP,$$ where the second equality follows from the fact that in an absolutely convergent series each term can be replaced by an infinite sum of terms with of the same sign, and the order of terms can be rearranged.
If $\xi$ is not necessarily simple, we find $\xi_n$ that satisfy properties (a)-(c). Then $$\lim\limits_{n \to \infty} \int_B \xi_n dP =\int_B \xi dP , \lim\limits_{n \to \infty} \int_{B_i} \xi_n dP =\int_{B_i} \xi dP.$$ Moreover,$$\lim\limits_{n \to \infty} \sum\limits_{i=1}^{\infty}\int_{B_i} \xi_n dP =\sum\limits_{i=1}^{\infty} \int_{B_i} \xi dP$$ since $\sum\limits_{i=1}^{\infty} |\int_{B_i} \xi_n dP-\int_{B_i} \xi dP|≤\sum\limits_{i=1}^{\infty} P(B_i) \sup\limits_{\omega \in B_i}|\xi_n(\omega)−\xi(\omega)| \le \sup\limits_{\omega \in B_i}|\xi_n(\omega)−\xi(\omega)| \to 0$ as $n \to \infty,$ where the first inequality is due to (2.2). Therefore, using (2.3) with $\xi$ replaced by $\xi_n$ (which is legal since $\xi_n$ are simple functions) and taking the limit as $n \to \infty,$ we obtain (2.3) for $\xi.$
If $\xi$ is a non-negative random variable with $E \xi < \infty,$ the $\sigma$-additivity of the integral implies that the function $ν(A) =\int_A \xi dP$ is itself a measure.
Remark 2.8. If, instead of the existence of $\int_B \xi dP,$ we assume that the integrals in the right-hand side of (2.3) exist and the series converges absolutely, then the integral in the left-hand side also exists. Not correct. This is easily justified by approximating $\xi$ with simple functions.
The reason I think the remark is incorrect is because $\lim\limits_{n \to \infty} \int_B \xi_n dP =\int_B \xi dP \, (*)$ does not mean anything unless $\int_B \xi dP$ exists. If you follow the remark, I agree that the proof for simple variables goes through, and that you still have $$\lim\limits_{n \to \infty} \sum\limits_{i=1}^{\infty}\int_{B_i} \xi_n dP =\sum\limits_{i=1}^{\infty} \int_{B_i} \xi dP,$$ and suspect the reason you can't copy the proof and take the limit at the end is because $(*)$ does not mean anything.
However, my reasoning might be incorrect. After all, by the same logic, $\lim\limits_{n \to \infty} \int_{B_i} \xi_n dP =\int_{B_i} \xi dP$ (complementary equation on the same line) doesn't mean anything in the original proof unless $\int_{B_i} \xi dP$ exists. However, I suspect we implicitly assume $\int_{B_i} \xi dP$ exists (hence the original proof uses the existence of both $\int_{B_i} \xi dP$ for all $i$ and $\int_B \xi dP$, while in the remark we only have the existence of $\int_{B_i} \xi dP$ and $\sum\limits_{i=1}^{\infty} \int_{B_i} \xi dP,$ which is not enough), which would alleviate this issue. In other words, the original proof uses existence of LHS and each term in the RHS to show the whole sum in the RHS exists and is equal to the LHS.
Regardless of whether my reasoning is right or wrong, I'd like to know if there's a concrete counterexample for which we can see $\int_B \xi dP$ clearly doesn't exist.