My actual question is:
Show that if $f$ is a uniformly continuous function on $\mathbb{R}^n$ and $f\in L^1(\mathbb{R}^n)$, then $f$ is bounded and $\displaystyle\lim_{\|\ x \|\ \to \infty} f(x)=0.$
Where the definition of $f\in L^1(\mathbb{R}^n)$ that I am working with is:
We say that $f\in L^1(\mathbb{R}^n)$ if there exists a sequence of simple step functions $(f_n)$ such that
(a) $\displaystyle\sum_{n=1}^{\infty} \displaystyle\int |f_n| < \infty$;
(b) $f(x)=\displaystyle\sum_{n=1}^{\infty} f_n(x) \hspace{.5cm} \forall x\in\mathbb{R}^n \space\ \text{such that} \space\ \sum_{n=1}^{\infty} |f_n(x)| < \infty.$
Then, the integral of $f$ is defined by $$\int f = \sum_{n=1}^{\infty} \int f_n.$$
What I have so far is:
Suppose $f\in L^1(\mathbb{R}^n)$ and is uniformly continuous on $\mathbb{R}^n$. We want to show that $f$ is bounded and \begin{align*} \displaystyle\lim_{\|\ x \|\ \to \infty} f(x)=0. \end{align*} Since $f\in L^1(\mathbb{R}^n)$ there exists a sequence of simple function $(f_j)$ such that $$f \simeq \sum_{j=1}^{\infty} f_j.$$ By way of contradiction, suppose that $$\lim_{\|\ x \|\ \to \infty} f(x) \neq 0.$$ Then, there would exist an $\epsilon>0$ and an increasing sequence of $\|\ x_n \|\ \to \infty$ with $|f(x_n)| > \epsilon$. In other words, $$\|\ x_1 \|\ \leq \|\ x_2 \|\ \leq \dots \|\ x_n \|\ \to \infty,$$ where we are taking subsequences, so we may suppose $\|\ x_{n+1} \|\ \geq \|\ x_n \|\ $. Then, by the uniform continuity of $f$ we have a $0<\delta <1$ such that for every $n$ and for every $x \in B_{\delta}(x_n)$, we have $|f(x)| \geq \frac{\epsilon}{2}$. Then, $$\int_\mathbb{R^{\text{n}}}|f(x)|dx\geq\sum_{n=1}^\infty\int_{B_{\delta}(x_n)}|f(x)|dx\geq\sum_{n=1}^\infty \delta\varepsilon=\infty,$$ a contradiction to $f \in L^1(\mathbb{R}^n)$. Therefore, \begin{align*} \displaystyle\lim_{\|\ x \|\ \to \infty} f(x)=0. \end{align*}
Now as far as boundedness of $f$, I thought it follows immediately because the condition \begin{align*} \displaystyle\lim_{\|\ x \|\ \to \infty} f(x)=0. \end{align*} automatically implies $f$ is bounded for $f\in C_0(\mathbb{R}^n) \cap L^1(\mathbb{R}^n) \implies f$ is bounded. Am I right? Note: $$C_0(\mathbb{R}^n) = \left\{ f:\mathbb{R}^n \to \mathbb{R} \big| \lim_{\|\ x \|\ \to \infty} |f(x)| = 0 \right\}.$$
If $f\in C_{0}$, then for a large $M>0$, we have $|f(x)|\leq 1$ for all $|x|\geq M$. Since $f$ is continuous on $\{|x|\leq M\}$, then $M:=\max_{|x|\leq M}|f(x)|<\infty$, so $\sup_{x\in{\bf{R}}^{n}}|f(x)|\leq M+1$.
If $f$ is not continuous, $f\in C_{0}\cap L^{1}$ does not imply that $f$ is bounded. Take for example that $f(0)=0$, $f(x)=1/|x|^{1/2}$ on $\{0<|x|\leq 1\}$, $f$ to be linear on $\{1\leq |x|\leq 2\}$ and $f(x)=0$ on $\{|x|\geq 2\}$ on one-dimensional.