If A $\subseteq$ E $\subseteq$ B, A and B are of infinite measure, is E measurable? I am aware of the proof to show that if A $\subseteq$ E $\subseteq$ B, where A
and B are both of finite measure, then if m(A) = m(B), E is measurable. 
However, is this the case too, if A and B are both of infinite measure?
 A: Certainly not; consider $A=(-\infty,0)$, $B=\Bbb R$.
The relevant condition is $m(B\setminus A)=0$. That's equivalent to $m(A)=m(B)$ if the sets have finite measure.
A: No: let $N$ be a nonmeasurable subset of $[0,1]$. Then take $A=(-\infty,0)$, $E=(-\infty,0)\cup N$, and $B=\Bbb R$. $E$ is not measurable since $N$ is not measurable.
A: I think that if $A \subseteq E \subseteq B$, then the fact  "$m(A)=m(B)<\infty \implies \mbox{E measurable}$" is only true for a complete measure space  (meaning a space where any subset of a measure 0 set is also measurable with measure 0).  See https://en.wikipedia.org/wiki/Complete_measure
So the interesting question to me seems to be why not define a "really complete" measure space so that any superset of a measure-$\infty$ set also has measure infinity.  I think the reason this is not done is that we want the intersection of two measurable sets to be measurable, but, given the examples in the other two answers, we could have the intersection of two measure-$\infty$ sets to be almost any crazy set we like.
In particular, if we start with Lebesgue measure on $\mathbb{R}$ and turn it into a "really complete" measure space on $\mathbb{R}$, then we can prove that any subset $C \subseteq [0,1]$ is measurable, since we can just take the intersection of $(-\infty, 0) \cup C$ with $[0,\infty)$:
$$ C = \underbrace{\{(-\infty, 0) \cup C\}}_{\mbox{really-measure $\infty$}} \cap \underbrace{[0,\infty)}_{\mbox{really-measure $\infty$}} $$  This is not a consequence we want, so we do not define "really complete" measure spaces.
