show that $f=\lim_{j \to \infty} f_j(x) \ \text{ a.e. everywhere}$

Consider the sequence of functions $\{f_n(x) \}$ , where $f_n(x)=\frac{\sin^n x}{x} , \ x \in \mathbb{R}$ such that $$1 \geq |f_1(x)| \geq |f_2(x)| \geq \cdots \geq \cdots \geq 0$$

$\text{Prove that$ \{f_n \}$converges pointwise almost everywhere to a Lebesgue integrable function$f$}$.

we will use the following Lemma called $\ \text{ Monotoncity Lemma }$ which is as follows:

If $f_j \in \mathcal{L}(\mathbb{R})$ is a monotone sequence, either $f_j(x) \geq f_{j+1}(x) \ \forall \ x \in \mathbb{R}$ and all $j \$ or $\ f_j(x) \leq f_{j+1}(x) \ \forall \ x \in \mathbb{R}$ and all $\ j$ and $\ \int f_j$ is bounded , then

$$f=\lim_{j \to \infty} f_j(x) \ \text{ a.e. everywhere}$$

But I can not implement this information to answer my question.

Whenever $|\sin(x)|<1$, $x\neq 0$, we have $\lim_{n\to\infty}f_n(x) = \frac{\sin^n(x)}{x} = 0$.
Set of points $x$ with property that $|\sin(x)|=1$ is countable. This means that the set of such points is of measure zero.
This means that $f_n$ converges to $0$ almost everywhere. $0$ is a Lebesgue integrable function. Hence the sequence $f_n$ converges to a Lebesgue integrable function almost everywhere.