$E$ is measurable, $m(E)< \infty$, and $f(x)=m[(E+x)\bigcap E]$ for all $x \in \mathbb{R}$ 
Question: $E$ is measurable, $m(E)< \infty$, and $f(x)=m[(E+x)\bigcap E]$ for all > $x \in \mathbb{R}$.  Prove $\lim_{x \rightarrow \infty} f(x)=0$.

First, since measure is translation invariant, I'm assuming that $(E+x)\bigcap E=E$.  But then I had this thought: if $E=\{1,2,3\}$ and $x=1$, then $E+x = \{2,3,4\}$.  So the intersection is just a single point.  This will have measure zero.
My question is, I'm not sure if this is the right train of thinking.  And, if it is, I'm not sure how to make this rigorous.
 A: Here's another direction to take my hint.
Setting as above $E_n=E\cap[-n,n]$ we have $E=\cup_{n=1}^\infty E_n$ and $\{E_n\}$ is
an increasing sequence, so $m(E)=\lim_{n\to\infty}m(E_n)$. In particular for any $\varepsilon$ we've got $n^*$ so that $$m(E\setminus E_{n^*})=m(E)-m(E_{n^*})<\varepsilon$$
Now $(E_{n^*}+2n^*) \cap E$ is contained in $E\setminus E_{n^*}$ and thus has measure less than $\varepsilon,$ while $(E\setminus E_{n^*} +2n^*)\cap E$ is contained in $E\setminus E_{n^*}+2n^*,$ which
also has measure at most $\varepsilon$ by translation invariance. 
So the whole set 
$$(E+2n^*)\cap E=[(E_{n^*}+2n^*) \cap E ]\cup [(E\setminus E_{n^*} +2n^*) \cap E]$$  has measure no more than $2\varepsilon$.
A: Well, it seems you are a bit confused about which object lives where..
By translation invariance, we indeed have $m(E)=m(E+x)$, but not $E=E+x$ as you wrote. Also, $\{1,2,3\}\cap\{2,3,4\}$ has two common elements, not just one:)
The hint in one of the comments to consider $E_n:=E\cap[-n,n]$ is a great idea, because in case $x>2n$, $E_n$ and $E_n+x$ will be disjoint, and
$$(E+x)\cap E = \bigcup_n \left( (E_n+x)\cap E_n \right) $$
so you can use continuity from below of $m$.
Denote $f_n(x):=m((E_n+x)\cap E_n)$. By the above argument, if $x>2n$, then $f_n(x)=0$.
We are looking for
$$\lim_{x\to\infty} m((E+x)\cap E) = \lim_{x\to\infty}\lim_{n\to\infty} f_n(x)$$
And exchange the limits.. ($f_n$ is nonnegative and bounded: $f_n(x)\le m(E)<\infty$)
