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 
$$|g(x)| \leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}} \int_x^{x+1} 1 dt$$
$$|g(x)|\leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}}$$
 A: 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,
$$
\lim_{n\to\infty}\int_n^{n+1}f(x)\,\mathrm{d}x=0
$$
A: Continuing from your last step, you have 
$$|g(x)|\leq\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}$$
where $\chi_{[x,x+1]}(t)$ is the indicator function on the interval $[x,x+1]$.
So $$\lim_{x\to\infty}|g(x)|\leq\lim_{x\to\infty}\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}$$
Since $|f(t)^2\chi_{[x,x+1]}|\leq|f(t)^2|$ which is integrable, by Lebesgue's Dominated Convergence Theorem, we can move the limit inside the integral, so 
$$\lim_{x\to\infty}\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}=\left(\int_\mathbb{R} \lim_{x\to\infty}f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}=0$$
since $\lim_{x\to\infty}\chi_{[x,x+1]}=0$
Hence $\lim_{x\to\infty}|g(x)|\leq 0$, and we can conclude from there.
A: 
The result is valid for functions in $C_0^\infty(\mathbb R)$. So, it is true for functions in $L^2(\mathbb R)$ (as well as for functions in $L^p(\mathbb R)$ with $1\leq p<\infty$) by density.

Details: As $C_0^\infty(\mathbb R)$ is dense in $L^2(\mathbb R)$, there is $(f_n)$ in $C_0^\infty(\mathbb R)$ such that 
$$\|f_n-f\|_{L^2(\mathbb R)}\xrightarrow{n\to\infty} 0.\tag{1}$$
Take $\varepsilon>0$. We want to show that there exist $M>0$ such that
$$x>M\quad\Longrightarrow \quad |g(x)|<\varepsilon.$$
From $(1)$ there is $n_0\in\mathbb N$ such that
$$\|f_{n_0}-f\|_{L^2(\mathbb R)}<\varepsilon.\tag{2}$$
As $f_{n_0}$ has compact support, there is $M>0$ such that
$$|x|>M\quad\Longrightarrow \quad|f_{n_0}(x)|=0.\tag{3}$$
From $(2)$, $(3)$ and your estimate,
$$\begin{align}
x>M\quad\Longrightarrow \quad |g(x)|\leq \|f\|_{L^2(x,x+1)}&\leq\|f_{n_0}-f\|_{L^2(x,x+1)}+\|f_{n_0}\|_{L^2(x,x+1)}\\
&\leq\|f_{n_0}-f\|_{L^2(\mathbb R)}+\|f_{n_0}\|_{L^2(x,x+1)}\\
&<\varepsilon+0=\varepsilon.
\end{align}$$
A: We have that $\int_{-\infty}^{+\infty}f(u)^2du<+\infty \Rightarrow \lim_{x\rightarrow +\infty} \int_{-\infty}^{x}f(u)^2du=\int_{-\infty}^{+\infty}f(u)^2du$ and so for $ε>0,\  \exists M>0\ :\ \forall x$ with $x>M$, we have:  
$|\int_{-\infty}^{x}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du |<\frac{ε^2}{2}\ (1)$
When $x>M$ we also have $x>M+1$. Thus $(1)$ and the triangle inequality yield that:
$\int_{x}^{x+1}f(u)^2du=|\int_{-\infty}^{x+1}f(u)^2du -\int_{-\infty}^{x}f(u)^2du|=\\ |(\int_{-\infty}^{x+1}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du )-(\int_{-\infty}^{x}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du )|\leq \\ |\int_{-\infty}^{x+1}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du |+|\int_{-\infty}^{x}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du |<\frac{ε^2}{2}+\frac{ε^2}{2}=ε^2$
Therefore for $ε>0,\  \exists M>0 : \forall x$ with $x>M$, we have:
${(\int_{x}^{x+1}f(u)^2du)}^{\frac{1}{2}}<ε \Rightarrow |g(x)|<ε$, by your attempt.
From this point the definition of the limit to infinity finishes the job.
A: You have a good start. To finish, use the folling result: If $g\in L^1(\mathbb R),$ then $\int_x^{x+1} |g| \to 0$ as $x\to \infty.$
Proof: Let $g_n = g\chi_{[n,n+1]}.$ Then $g_n \to 0$ pointwise everywhere, and $|g_n| \le |g|$ on $\mathbb R.$ By DCT, $\int_{\mathbb R} |g_n| \to 0.$ That proves the result for $x= 1,2,\dots,$ and that will give the result for arbitrary $x\to \infty.$
In your problem, use $g=f^2.$
A: $d\mu= f^2 dt$ defines a positive finite measure on $\mathbb{R}$ (this is just another way of writing $\mu (A) = \int_{A} f(t)^2 dt$ if it's not clear), since $f$ is square-integrable.
Then, $\mu(x,x+1)= \mu(-\infty,x+1) - \mu(-\infty,x)$ (this is allowed since both sets have finite measure).
Then $$ \lim_{x \to +\infty} \int_{x}^{x+1} f(t)^2 dt = \lim_{x \to +\infty} \mu(x,x+1) = \mu(\mathbb{R}) - \mu(\mathbb{R}) = 0$$
