Let $f:[a,\infty)\rightarrow \mathbb{R}$ be a uniformly continuous function. $\int_{a}^{\infty} f$ converges.Prove that $\lim_{x\to\infty} f(x)=0$ 
Let $f:[a,\infty)\rightarrow \mathbb{R}$ be a uniformly continuous function in that range. $\int_{a}^{\infty} f$ converges. Prove that $\lim_{x\to\infty} f(x)=0$
Hint: Use the sequence $F_n(x)=n\int_{x}^{x+\frac{1}{n}} f$.

Honestly I have been trying to solve this one for some time but the hint really confuses me.
I have tried to mess around with $F_n(x)$ a bit, for example by using the fundamental theorem but it still seems like such a random choice and I can't make anything out of it.
Any guidance/explanations will be appreciated.
Please use the hint in the question.

First of all, note that for all $n\in\mathbb{N}$
$$\lim_{x\rightarrow\infty}F_n(x)=n\left[\lim_{x\rightarrow\infty}\left(\int_a^{x+\frac{1}{n}}f(t)\mathrm{d}t- \int_a^xf(t)\mathrm{d}t\right)\right]=n\left[\int_a^{\infty}f(t)\mathrm{d}t-\int_a^{\infty}f(t)\mathrm{d}t\right]=0.$$
Now let $\varepsilon>0$ be arbitrary. By uniform continuity, there is a $\delta>0$ such that for all $t,x\in\left[a,\infty\right)$, we have $|f(t)-f(x)|<\varepsilon$ whenever $|t-x|<\delta$. Pick $N\in\mathbb{N}$ such that $\frac{1}{n}<\delta$ for all $n\ge N$. Then for all $x\in\left[a,\infty\right)$, we have
$$\left|n\int_x^{x+\frac{1}{n}}f(t)\mathrm{d}t-f(x)\right|=\left|n\int_x^{x+\frac{1}{n}}f(t)-f(x)\mathrm{d}t\right|\le n\int_x^{x+\frac{1}{n}}|f(t)-f(x)|\mathrm{d}t\le\varepsilon.$$
Since $\varepsilon$ and $x$ were arbitrary, we conclude that $\lVert F_n-f\rVert\rightarrow0$ as $n\rightarrow\infty$, i.e. $F_n\rightarrow f$ uniformly. Thus,
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\lim_{n\rightarrow\infty}F_n(x)=\lim_{n\rightarrow\infty}\lim_{x\rightarrow\infty}F_n(x)=0.$$
