Showing $\int_x^{x+1}f(t)\,dt \xrightarrow{x\to\infty}0$ for $f\in L^2 (\mathbb{R})$ 
Show that if $f\in L^2 (\mathbb{R})$ then 
  $$\lim_{x \rightarrow \infty} g(x)=0,$$
  where 
  $$g(x)=\int_x^{x+1}f(t)\,dt$$

Since $f \in  L^2 (\mathbb{R})$ then $f \in L^2[x,x+1]$ for every $x$. Then 
$$|g(x)| \leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}} \int_x^{x+1} 1 dt$$
$$|g(x)|\leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}}$$

Using Jensen's Inequality, the sum
$$
\begin{align}
\sum_{n=0}^\infty\left(\int_n^{n+1}|f(x)|\,\mathrm{d}x\right)^2
&\le\sum_{n=0}^\infty\int_n^{n+1}|f(x)|^2\,\mathrm{d}x\\
&=\int_0^\infty|f(x)|^2\,\mathrm{d}x\\
&\le\|f\|_{L^2(\mathbb{R})}^2
\end{align}
$$
converges, which means the terms must go to $0$. That is,
$$
\lim_{n\to\infty}\int_n^{n+1}f(x)\,\mathrm{d}x=0
$$
