Is my attempt at proving this corollary of the LDCT correct?

Question

$\mathbf{LDC\ Theorem:}$ Suppose that $(f_{n})$ is a sequence of integrable functions, $f_{n}: X \rightarrow \overline{\mathbb{R}}, $ on a measure space $(X,\mathcal{A},\mu)$ that converges pointwise to a limiting function $f:X \rightarrow \overline{\mathbb{\mathbb{R}}}.$ If there is an integrable function $g:X \rightarrow [0,\infty]$ such that $$|f_{n}(x)\leq g(x)| \qquad \text{ for all } x\in X \text{ and } n\in \mathbb{N}.$$ then $f$ is integrable and $$\lim_{n \rightarrow \infty} \int f_{n} d\mu = \int f\ d\mu.$$

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$\mathbf{Corollary:}$ Let the mapping $x \rightarrow f(x,\tau)$ be a measurable function on $X$ for each $\tau \in [a,b],$ and suppose that for some $t \in [a,b]$ $$f(x,t)=\lim_{\tau \rightarrow t} f(x,\tau) \text{ for each } x \in X.$$ If there exists a function $g\in L^{1}(X,\mathcal{M}, \mu)$ such that ${f(x,\tau)} \leq g(x)$ for all $x \in X\ \tau \in [a,b]$, prove that $$\int f(x,t)\ d\mu(x)=\lim_{\tau \rightarrow t} \int f(x,\tau)\ d\mu(x).$$

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$\textbf{Attempt}:$ (Heavily Edited)

Let $F(x):= f(x,\tau)$ be a measurable function on $X$, for $t\in [a,b]$ $f(x,t)=\lim\limits_{\tau \rightarrow t} f(x,\tau)$ and $g\in L^{1}(X,\mathcal{M}, \mu)$ such that $|{f(x,\tau)}| \leq g(x)$ for each $x \in X$. Since $f(x,t)=\lim\limits_{\tau \rightarrow t}\ f(x,\tau)$ by SFCL(as hinted by Calvin) for any $(\tau_{n}) \subset [a,b]$ such that $(\tau_{n}) \rightarrow t$ then $\lim\limits_{n \rightarrow \infty}$ $f(\tau_{n})=f(t).$ Since we have a sequence of integrable functions $F_{n}(x):=f(x,\tau_{n}) \rightarrow f(x,t)$ that converge pointwise and $|{f(x,\tau)}| \leq g(x)$ by LDCT for all ${\tau_{n}\rightarrow} \subset [a,b]$ such that $\tau_{n} \rightarrow t$ then: $$ \lim_{n \rightarrow \infty} \int f(x, \tau_{n}) d\mu = \int f (x, \tau)\ d\mu \Longleftrightarrow \lim_{\tau \rightarrow t} \int f(x,\tau)\ d\mu(x) = \int f(x,t)\ d\mu(x). \Box $$

Answer 1

Summary of comments before it gets too long- You need to remember:

Definition. A sequence of elements in $X$ is a map $a:\mathbb N \to X$, usually written for $n\in\mathbb N$ as $a_n$ instead of $a(n)$.

And

Claim. The set $[a,b]$ is not the set of values of any one sequence $a_n$ of elements of $\mathbb R$.

Proof. The set $\{ a_n : n\in\mathbb N\}$ is countable, while $[a,b]$ is uncountable.

To prove the main result in the question, I strongly suggest you remind yourself of the following

Claim. (Sequential formulation of continuous limit) For a function $f:[a,b]\to Y$, the continuous limit $\lim_{x\to c} f(x) = L\in Y$ holds iff for any sequence $(x_n)\subset [a,b]$ with the sequence limit $\lim_{n\to\infty} x_n = c$, we have the sequence limit $\lim_{n\to\infty} f(x_n) = L$.

Note carefully that sequence limits and continuous limits have different definitions.