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

Question

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 $f$ }$.

Answer:

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.

Please help me

Answer 1

We don't need this theorem.

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.