I've been working on this problem whilst studying for a comprehensive exam, and I've come up with a solution that I don't like. I'd appreciate some critiques. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a uniformly continuous function with the property that $f\in L^{1}\left(\mathbb{R}\right)$. The claim is that $f$ is necessarily bounded on $\mathbb{R}$. My solution that I don't like is as follows:
Suppose to the contrary that $f$ fails to be bounded on $\mathbb{R}$. Let $M\in\mathbb{R}^+$ be given, and find $x_0\in\mathbb{R}$ satisfying $\left|f\left(x_0\right)\right|>M$. Because $f$ is, by assumption, continuous, there is a $\delta>0$ such that for all $x\in\left(x_0-\delta,x_0+\delta\right)$, we have $\left|f(x)\right|>M$. It follows that $$\int_{\mathbb{R}}{|f|\hspace{3pt}dm}\geq\int_{\left(x_0-\delta,x_0+\delta\right)}{|f|\hspace{3pt}dm}\geq\int_{\left(x_0-\delta,x_0+\delta\right)}{M\hspace{3pt}dm}=M\cdot 2\delta.$$ But $M$ was arbitrary, so we have that, essentially, $$\int_{\mathbb{R}}{|f|\hspace{3pt}dm}\geq M$$ for any $M\in\mathbb{R}^+$, and therefore $$\int_{\mathbb{R}}{|f|\hspace{3pt}dm}=\infty.$$ Hence $f\notin L^{1}(\mathbb{R})$, a contradiction. Therefore it must be the case that our assumption that $f$ was unbounded is false. $\therefore f$ is bounded, as claimed.
Primarily, I don't like the fact that I seemed not to use $uniform$ continuity of $f$ in the proof. I'd appreciate any thoughts, critiques, etc.
Why not show even more: Actually, $f\in C_0(\mathbb{R})$, that is, $\lim_{x\to-\infty}|f(x)|=\lim_{x\to+\infty}|f(x)|=0$ (we say that $f$ vanishes at $\infty$). This implies that $f$ is bounded.
Let's treat the case $x\to+\infty$. If we didn't have $\lim_{x\to+\infty}|f(x)|=0$, there would exist $\varepsilon>0$ and an increasing sequence $x_n\to\infty$ with $|f(x_n)|>\varepsilon$. Taking subsequences, assume $x_{n+1}>x_n$. Uniform continuity gives us $\delta<1$ such that for every $n$ and for every $x\in (x_n-\delta,x_n+\delta)$, we have $|f(x)|\geq\varepsilon/2$. Then $$\int_\mathbb{R}|f(x)|dx\geq\sum_{n=1}^\infty\int_{x_n-\delta}^{x_n+\delta}|f(x)|dx\geq\sum_{n=1}^\infty \delta\varepsilon=\infty,$$ a contradiction.