First moment of the measure and interval escaping to infinity Let $\mu$ be a finite Borel measure on $\mathbb{R}$, and 
$$L = \lim_{t \to +\infty} \int_{t-1}^{t+1} x d \mu (x).$$
If $\displaystyle  \int_\mathbb{R} |x| d \mu (x) < \infty$, then by Dominated Convergence $L<\infty$. 
What would be an example of $\mu$ for which $L=\infty$? I tried
$$d \mu (x) = \frac{d x}{(1+|x|)^p},$$
then the conditions $\mu$ being finite and its first moment infinite imply $p \in (1,2]$, however for this $L<\infty$. 
 A: There's no such $\mu$.  Indeed, if the limit $L$ exists it must equal 0.
Suppose to the contrary that $L>0$.  Set $F(t) = \int_{t-1}^{t+1} x\,d\mu$; then there exists an integer $N>0$ such that $F(t) \ge L/2$ for all $t \ge N$.  Then for such $t$ we have
$$\frac{L}{2} \le \int_{t-1}^{t+1} x\,d\mu \le (t+1) \mu([t-1, t+1])$$
since $x \le t+1$ on the interval in question.  Rearranging, we have $\mu([t-1, t+1]) \ge \frac{L}{2(t+1)}$.  Then
$$\begin{align*}\mu(\mathbb{R}) &\ge \mu([N-1, N+1]) + \mu([N+2, N+4]) + \mu([N+5, N+7]) + \cdots \\ &\ge \frac{L}{2}\left(\frac{1}{N+1} + \frac{1}{N+4} + \frac{1}{N+7} + \cdots\right) \\ &= \infty\end{align*}$$
because the harmonic series diverges.  This contradicts the assumption that $\mu$ was finite.  
In fact, this shows that in general $\liminf_{t \to +\infty} \int_{t-1}^{t+1} x\,d\mu = 0$.  It is of course possible that the $\limsup$ is nonzero or infinite.  Consider for instance the measure $\mu$ that assigns measure $2^{-n}$ to the point $2^{2n}$.
