Can the integration of integrable sections of a measurable function of two variables ever result in a non-measurable function?

EDIT: The following only provides a partial answer, since it is not clear at all that the first property of the question ($f(x, \cdot) \in L^1(Y, \mathcal{T}, \nu)$) is fulfilled for the given example.

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Let $(X, \mathcal{S})$ and $(Y, \mathcal{T})$ both be the real line with the Borel sigma algebra. Note that the product sigma algebra is again the Borel sigma algebra (but on $\Bbb{R}^2$).

It is well-known (see Projection of Borel set from $R^2$ to $R^1$) that not every projection of a Borel set is a Borel set. Hence, let $M \subset \Bbb{R}^2$ be a Borel set such that the projection $\pi_1 (M)$ is not Borel measurable.

Let $\mu$ be the counting measure on the real line. If $$ F(x) := \int_{\Bbb{R}} 1_M (x,y) d \mu(y) $$ was measurable, then so would be the set $$ \pi_1 (M) = \{x \,:\, \exists y : (x,y) \in M\} = \{x \,:\, F(x) > 0\}. $$

Let $\mathcal{B}$ denote the class of Borel subsets of $[0,1]$ and let $A \subseteq [0, 1]$ be a non Borel set. Let $f$ be the characteristic function of the graph of a bijection from $A$ to $[0, 1]$. Then $f$ is $\mathcal{B} \otimes \mathcal{P}([0, 1])$-measurable (check) and the map $x \mapsto \int f(x, y) d\mu(y)$ is non-zero precisely on $A$ where $\mu$ is counting measure.

Edit: I was asked to provide more details so here they are.

Suppose $W \subseteq \mathbb{R}^2$ is such that every horizontal section $W^y = \{x : (x, y) \in W\}$ is closed. Then $W \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R})$. To see this, define, for each interval $J$ with rational endpoints, $Y_J = \{y : W^y \cap J = \phi\}$. As each $W^y$ is closed, we have $$ W = \mathbb{R}^2 \Big\backslash \bigcup \{J \times Y_J : J \text{ is an interval with rational endpoints}\} \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R}). $$ It follows that the graph of every partial bijection on $\mathbb{R}$ is in $\mathcal{B} \otimes \mathcal{P}(\mathbb{R})$.

Now choose any non Borel set $A \subseteq \mathbb{R}$ and an injection $i: A \to \mathbb{R}$. Define $f: \mathbb{R}^2 \to \mathbb{R}$ to be the characteristic function of the graph of $i$. Let $\mu$ be the counting measure on $\mathbb{R}$. The map $x \mapsto \int f(x, y) d\mu(y)$ is precisely the characteristic function of $A$ and hence is non-Borel.