Extending the Dominated Convergence Theorem to Continuous Limits

I understand the usual dominated convergence theorem is something as follows:

Let $\{f_n\}_{n\in\mathbb{N}}$ be a sequence of real-valued measurable functions on a measure space $(\mathscr{X},\mathcal{A},\mu)$. Moreover suppose that $f_n\rightarrow f$ in measure and $\left\vert f_n(x)\right\vert \le g(x)$ for all $x\in\mathscr{X}$ and $g$ is integrable. Then $\int f_n \rightarrow \int f$.

I now want to say/generalise this to:

1) If $f:\mathscr{Y}\times \mathscr{X}\longrightarrow \mathbb{R}$ such that \begin{equation*} \mu\left\{x\in\mathscr{X} : \lim_{y\rightarrow y_0} f(y,x) \ne f(y_0, x)\right\} =0 \end{equation*} and there exists a function $g:\mathscr{Y}\times \mathscr{X}\longrightarrow \mathbb{R}$ such that there exists $\tilde y\in\mathscr{Y}$ for which: \begin{equation*} \sup_{y\in \mathscr{Y}} \left\vert f(y, x)\right\vert \le g(\tilde y,x) \qquad \forall x\in\mathscr{X} \end{equation*} Then \begin{equation*} \lim_{y\rightarrow y_o} \int_{\mathscr{X}} f(y, x)d\mu = \int_{\mathscr{X}}f(y_0,x)d\mu \end{equation*}

2) And I want to even further generalise this to: For each $\epsilon>0$ \begin{equation*} \lim_{y\rightarrow y_o}\mu\left\{x\in\mathscr{X} : \left\vert f(y,x) - f(y_0, x)\right\vert >\epsilon\right\} =0 \end{equation*} and there exists a function $g:\mathscr{Y}\times \mathscr{X}\longrightarrow \mathbb{R}$ such that there exists $\tilde y\in\mathscr{Y}$ for which: \begin{equation*} \sup_{y\in \mathscr{Y}} \left\vert f(y, x)\right\vert \le g(\tilde y,x) \qquad \forall x\in\mathscr{X} \end{equation*} Then \begin{equation*} \lim_{y\rightarrow y_o} \int_{\mathscr{X}} f(y, x)d\mu = \int_{\mathscr{X}}f(y_0,x)d\mu \end{equation*}

In saying this I have spent the last two afternoons trying to prove these statements to myself, and I am not very confident in my proofs. Could someone please provide a rigorous proof of these statements as I could definitely benefit from learning how other's have approached the problem as well? Many thanks.

• What do you mean by $y \to y_0$? Your measure space also carries a topological structure? – Vinícius Novelli Mar 8 '18 at 1:27
• What do you mean by $$\begin{equation*} \sup_{y\in \mathscr{Y}} \left\vert f(y, x)\right\vert \le g(y,x) \qquad \forall x\in\mathscr{X} \end{equation*}$$ The right side has a free $y$. – Robert Israel Mar 8 '18 at 1:43
• @ViníciusNovelli. Sorry I was not specific enough - the specific example I actually want to apply this question/theorem to is where $\mathscr{Y}$ and $\mathscr{X}$ are euclidean spaces with the usual euclidean norm/metric. However I was hoping that in this question I could just have $\mathscr{X}$ and $\mathscr{Y}$ as general metric spaces with respective metrics $d_X$ and $d_Y$...So when I say $y\rightarrow y_0$ then I would mean this in the usual sense of pointwise convergence of $y$ to $y_0$ in a metric space. – User086688 Mar 8 '18 at 5:04
• @RobertIsreal. Ah very good point. With this statement I am just trying to capture that somehow $f(y, x)$ is "dominated". Perhaps I should reword this as $\exists$ $\tilde y\in\mathscr{Y}$ such that: \begin{equation*} \sup_{y\in\mathscr{Y}}\left\vert f(y,x)\right\vert \le g(\tilde y, x) \quad \forall x\in\mathscr{X} \end{equation*} I will edit the question accordingly now - thank you for pointing this out! – User086688 Mar 8 '18 at 5:08

The statement $\lim_{y \to y_0} \int f(y,x) d\mu(x) = \int f(y_0,x) d\mu(x)$ is equivalent to the statement $\lim_{n \to \infty} \int f(y_n,x) d\mu(x) = \int f(y_0,x) d\mu(x)$ for every sequence $y_n \to y_0$. So I don't see anything new here. It should be noted, however, that measurability of the limit is an essential item in the hypothesis: a limit through continuum of measurable functioins is not automatically measurable.