Q:

Showing the Continuity of a Lebesgue Integral Function over an Uncountable Domain

$\fbox{Hypothesis}$
Suppose we have a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ that gives rise to two families of single-variable functions:
$\{f(x,t)$ for fixed $t\in \mathbb{R}\} = \{h_x\}_{x \in \mathbb{R}}$ s.t. $h_x: \mathbb{R} \rightarrow \mathbb{R}$ and is continuous
$\{f(x,t)$ for fixed $x\in \mathbb{R}\} = \{g_t\}_{t \in \mathbb{R}}$ s.t. $g_t: \mathbb{R} \rightarrow \mathbb{R}$ and is Lebesgue integrable
Now suppose further that $\exists g: \mathbb{R} \rightarrow \mathbb{R}$ s.t. $g$ is Lebesgue integrable and $\forall t \in (a,b)$ we have that $|g_t| \le g$.
Now let $F(t) = \int_\mathbb{R} g_t$.
Then I want to show that $F$ is continuous which is true if and only if $\lim_{t_n\to t_0} \int g_{t_n} = \int g_{t_0}$.

A:

$\fbox{Constructive Proof}$
(1) Let $\{t_n\}$ be an arbitrary real sequence s.t. $t_n \rightarrow t_0$ with $t_n \ne t_0$.
(2) Then $\{g_{t_n}\}$ converges to $g_{t_0}$ pointwise since $\forall x \in \mathbb{R}$ we have that
$$
\lim_{n\to\infty} g_{t_n}(x) = \lim_{n\to\infty} h_x(t_n) = h_x(t_0) = g_{t_0}(x).
$$
(3) Now consider that $|g_{t_n}| \le g$ for great enough $n$ since $t_n \rightarrow t_0 \in (a,b)$ whereby $|g_{t_n}| \le g$ by hypothesis.
(4) Then applying The Dominated Convergence Theorem to $\{g_{t_n}\}$ we obtain that
$$
\int g_{t_0} = \lim_{n\to\infty} \int g_{t_n}
$$
and since $\{t_n\}$ was an arbitrary sequence, it follows that more generally
$$
\lim_{t_n\to t_0} \int g_{t_n} = \int g_{t_0}
$$
as desired for $F$ to be continuous at $t_0 \in (a,b)$.

