Monotone convergence theorem for a generic $f(x,y)$ instead of $f_n(x)$ I was wondering if it is safe to say that the monotone convergence theorem
$$\lim_{n\to\infty} \int_X f_n(x) =  \int_X \lim_{n\to\infty} f_n(x)$$
(where $f_n(x)$ is a non-decreasing sequence, etc.) is applicable even if one has a generic $f_y(x)=f(x,y)$, so considering a real parameter $y$ instead of a sequence of functions $f_n$.
Is it correct to say that if the dependence on $y$ is monotonic then
$$\lim_{y\to y_0} \int_X f(x,y) =  \int_X \lim_{y\to y_0} f(x,y)\ \ \ ?$$
Perhaps I should simply not refer to the monotone convergence theorem.
Thanks and sorry for being a little approximate in describing the problem and in the notation.
 A: We claim as follows:

Claim:
$f(x,y)^\rightarrow_\rightarrow \varphi(x), \ \ y\to y_0, ~x \in X \tag{1}$ if and only if $\forall \{y_n\}: y_n \to y_0 \implies f_n(x) = f(x,y_n)^\rightarrow_\rightarrow \varphi(x) \tag{2}$

Proof: $(1) \implies (2)$ is quite easy to see.
Assume $(2)$ is true. That is, $\forall \varepsilon>0, \exists \delta > 0, \exists n_0\in \mathbb N, \forall n > n_0:$ $$|y_n-y_0| < \delta \implies |f(x,y_n) - \varphi(x)| < \varepsilon \tag{$\forall x \in X$}$$
Now, suppose $(1)$ is false. That is, $\exists \varepsilon > 0, \forall \delta > 0, \exists y (0<|y-y_0|<\delta), \exists x:$
$$|f(x,y) - \varphi(x)| \ge \varepsilon$$
However, this implies for any sequence $\{\delta_n\}: \delta_n \to 0$, we have

*

*$\exists y_1(|y_1-y_0|<\delta_1) \implies |f(x, y_1) - \varphi(x)|\ge\varepsilon$

*$\exists y_2(|y_2-y_0|<\delta_2) \implies |f(x, y_2) - \varphi(x)|\ge\varepsilon$

*$\dots$

*$\exists y_n(|y_n-y_0|<\delta_n) \implies |f(x, y_n) - \varphi(x)|\ge\varepsilon$
and this contradicts the initial assumption.

Now, you can apply this result to your problem.
