Show that if $f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous such that
$$\int_{-\infty}^{\infty}|f(x)| < \infty \Rightarrow \lim_{|x| \to \infty}|f(x)| = 0$$
I already try this but I'm not sure it works:
Consider
$$A_n = \{x \in \mathbb{R}: |f(x)| \geq n \}$$
Then
$$m(A_n)n \leq \int_{A_n}|f(x)| \leq \int_{\mathbb{R}}|f(x)| < \infty$$
Hence,
$$m(A_n) \leq \frac{1}{n}\int_{\mathbb{R}}|f(x)| < \infty$$
Thus,
$$\lim_{n\to \infty}m(A_n) = 0 = m(\cap_{n = 1}^{\infty}A_n)$$
Therefore, $m(\{x \in \mathbb{R}: |f(x)| = \infty\}) = 0$
Did this implies that $f$ goes to 0 as $|x| \to \infty$? I don't think so. I was wondering if I can use another approach using the fact that it is lebesgue integrable and uniformly continuous. Then the integral of $f$ is the near the integral of a simple function that is bounded and vanish outside a set of finite measure, but I do not know how to attack this problem using this approach.
Suppose that $f$ is uniformly continuous but that $\lim_{|x| \to \infty} f(x) \neq 0$.
Then there exists some $\alpha>0$ and a sequence $x_n \to \infty$ (or possibly $x_n \to -\infty$) such that $|f(x_n)|>\alpha$. Wlog we may assume that $|x_n-x_m|>1$ for $m \neq n$.
By uniform continuity, there exists some $\delta>0$ such that for every $n$, we have $|f(x)|>\alpha/2$ whenever $|x_n-x|< \delta$.
Consequently $$|f| \geq \alpha/2 \sum_n 1_{[x_n-\delta,x_n+\delta]}$$ So what can you conclude?