Let, $p \in [1,\infty)$. Suppose, $g \in L^p(\mathbb{R}^{n},m)$ where $m$ is the n'th dimensional Leb. measure, and $g$ is uniformly cont. Then, $$\lim_{||x|| \rightarrow \infty}g(x) = 0.$$
Suppose, $\lim_{||x|| \rightarrow \infty}g(x) \ne 0$. Then, $\exists \epsilon >0:$ $$\forall M \in \mathbb{R}, \exists \alpha, ~ ||\alpha|| > M ~ \land ~ |g(\alpha)| > \epsilon.$$ Then we can construct an increasing sequence (w.r.t. the norm) $\{\alpha_n\}_{n \ge 1}$ such that, $|g(\alpha_n)| > \epsilon$ and $||x_n|| < ||x_{n+1}|| + 1$ for all $n \in \mathbb{N}$.
By uniform continuity we have that, $\exists \delta,\forall x,y \in \mathbb{R}^{n}$: $$||x-y|| < \delta \implies |f(x) - f(y)| < \frac{\epsilon}{2}.$$ Let, $S = \bigcup_{n \ge 1} B(\alpha_n,\min\{\delta,1\}).$ The ball of radius $\min\{\delta,1\}$ centered at $\alpha_n$. We have that,
$$\int_{S}(\frac{\epsilon}{2})^{p} ~ dm \le \int_{\mathbb{R}^n}|g|^p ~ dm.$$
Observe that, $\mu(S) = \sum^{\infty}_{n=1} \min\{1,\alpha\} = \infty.$ Thus, $$\infty \le \int_{\mathbb{R}^n}|g|^p ~ dm. $$
Which is a contradiction to $g \in L^{p}(\mathbb{R}^{n},m)$.
EDIT: Thank you, @MathematicsStudent1122 for the tip.