Proving a corollary of Lebesgue's Dominated Convergence Theorem
In reading Stein & Shakarchi's Real Analysis, I noticed that the authors apply the Dominated Convergence Theorem to not only sequences of functions $\{f_n\}_{n=1}^{\infty}$, but also to more general families of functions of the form $\{f_{\delta}\}_{\delta > 0}$. In trying to understand this application of the DCT here, I've come up with the following "Theorem" and "Proof" that justifies it's usage in this way.
My question: Is the following "Theorem" and "Proof" correct? I want to make sure my logic sound.
Theorem: Let $h: \mathbb{R}^{d_1} \times \mathbb{R}^{d_2} \rightarrow \mathbb{C}$ be a measurable function and let $y' \in \mathbb{R}^{d_2}$. Suppose there is some function $g \in L^1(\mathbb{R}^{d_1})$ such that $|h(x,y)| \leq g(x)$ for a.e. $(x,y) \in \mathbb{R}^{d_1} \times \mathbb{R}^{d_2}$, and that $h$ is continuous at $(x,y')$ for all $x \in \mathbb{R}^{d_1}$. Then $$ \lim_{y \to y'} \int_{\mathbb{R}^{d_1}} h(x,y) \,dx = \int_{\mathbb{R}^{d_1}} h(x,y') \,dx. $$
Proof: Since $h$ is measurable on $\mathbb{R}^{d_1} \times \mathbb{R}^{d_2}$, it follows by Tonelli's theorem that $h^y(x) = h(x,y)$ is measurable on $\mathbb{R}^{d_1}$ for a.e. $y \in \mathbb{R}^{d_2}$. Therefore, we may select a sequence $\{y_n\}_{n=1}^{\infty}$ in $\mathbb{R}^{d_2}$ such that $\lim\limits_{n \to \infty} y_n = y'$ and $h^{y_n}(x) = h(x,y_n)$ is measurable for all $n$ [see footnote]. Clearly we have $|h^{y_n}(x)| \leq g(x)$ for a.e. $x$ and all $n \geq 1$. Also, because $h$ is continuous at the points $(x,y')$, we have $\lim\limits_{n \to \infty} h^{y_n}(x) = h(x,y')$. Therefore,
\begin{align*}
\lim_{y \to y'} \int_{\mathbb{R}^{d_1}} h(x,y) \,dx &= \lim_{n \to \infty} \int_{\mathbb{R}^{d_1}} h^{y_n}(x) \,dx && \\[5pt]
&= \int_{\mathbb{R}^{d_1}} h(x,y') \,dx, && (\text{Dominated Convergence Theorem})
\end{align*}
where the second to last equality is by the fact that $\lim\limits_{x \to a} f(x) = L \iff \lim\limits_{n \to \infty} f(x_n) = L \text{ whenever } \lim\limits_{n \to \infty} x_n = a$.
My question: Does the theorem and conclusion seem sound? I just want to make sure, as I haven't seen this result stated explicitly in the text. Any further comments on the theorem/proof are welcome.
Footnote (Justification for why we can find such a sequence $\{y_n\}_{n=1}^{\infty}$): By Tonelli's Theorem, $h^{y}(x)$ is measurable for a.e. $y \in \mathbb{R}^{d_2}$. For each $n \geq 1$, the ball $B_{1/n}(y')$ in $\mathbb{R}^{d_2}$ has positive measure, and therefore contains some $y_n$ such that $h^{y_n}$ is measurable. Picking $y_n \in B_{1/n}(y')$ for each $n$, we then have $|y_n - y'| \leq 1/n$ for all $n \geq 1$, and hence $\lim_{n \to \infty} y_n = y'$, as desired.
As long as $\delta$ lives in a metric space, the dominated convergence theorem still works since convergence in a metric space is equivalent to convergence along all sequences.
For example: If $f_{\delta} \to f$ pointwise as $\delta \to 0$, then $f_{\delta_n} \to f$ pointwise for any sequence $\delta_n \neq 0$, $\delta_n \to 0$. So for any sequence $\delta_n \neq 0$, $\delta_n \to 0$ the dominated convergence theorem gives $$\lim_{n \to \infty}\int f_{\delta_n} = \int f.$$ Since this holds for all sequences, $$\lim_{\delta \to 0}\int f_{\delta} = \int f.$$