Let $(X,\mathcal X,\mu)$ and $(Y,\mathcal Y,\nu)$ be finite measure spaces. We denote the product $\sigma$-algebra by $\mathcal X \otimes \mathcal Y$. Let $\{ f_n \}$ be a sequence of functions where each $f_n\colon X\times Y \to \mathbb R$ is $\mathcal X \otimes \mathcal Y$-measurable.
Suppose that for each $y\in Y$, $f_n(\cdot,y)$ converges in $\mu$-measure to $f(\cdot,y)$. We may also assume that $\vert f_n(x,y)\vert \le G(x,y)$, where $G$ is $\nu$-integrable for $\mu$-almost all $x$. Do we then have that, for some appropriate notion of convergence, $$ \lim_{n\to \infty} \int_Y f_n (\cdot,y) \, \mathrm d\nu = \int_Y f(\cdot,y)\,\mathrm d \nu? \tag{1}\label{1} $$
By convergence in $\mu$-measure for each $y$, I mean that, for each $\varepsilon > 0$ and each $y \in Y$, $$ \lim_{n\to \infty} \mu \left( \left\{ x: \vert f_n (x,y) - f(x,y) \vert > \varepsilon \right\} \right) = 0. $$
Some Thoughts
If for each $y$, $f_n(x,y)\to f(x,y)$ for $\mu$-almost all $x$, then \eqref{1} holds, where the limit is taken for $\mu$-almost all $x$. This is an easy consequence of the Dominated Convergence Theorem.
If $f_n(\cdot,y)\to f(\cdot,y)$ in $\mu$-measure for each $y$, then one might consider passing to a subsequence that converges for $\mu$-almost all $x$. However, this subsequence now depends on $y$, so it is not clear how to apply the previous argument to this case.
If instead $f_n(\cdot,y) \to f(\cdot,y)$ in $\mu$-measure uniformly in $y$ (that is, $\sup_y \vert f_n(\cdot,y) - f(\cdot,y) \vert \to 0$ in $\mu$-measure), then \eqref{1} also holds, taking limits in $\mu$-measure. To see this, note that $$ \mu \left( \left\{ x: \left\vert \int_Y (f_n (x,y) - f(x,y))\,\mathrm d \nu \right\vert > \varepsilon \right\} \right) \le \mu \left( \left\{ x: \sup_y \left\vert f_n (x,y) - f(x,y)\right\vert\ > \frac{\varepsilon}{\nu(Y)} \right\} \right) $$ where the right-hand side goes to $0$ as $n \to \infty$. (The finiteness of $\nu$ is important here.)
Does anyone have any suggestions for the general case?