Sequence of Measurable Sets Suppose that $ E $ is a  measurable subset of $ R  $.   Suppose $    \left    \{    E_{i}    \right \}  $ is sequence of measurable subsets of $ E $. For any $ x  \in  E $, there exist an $ N_{x} $ such that for all $ n >  N_{x}    $, we have $ x   \notin  E_{n} $. Does it follow that $       \lim _ {   n      \to   \infty   }   m (  E_{ n } )  =  0  $? If not, under what conditions, can we make it happen?     
 A: Take the semi-open intervals $[n, n+1)$ as subsets of $E = \mathbb R$ as $n\to\infty$. Their measure is constantly $1$ so is the limit. But every single real is contained in at most one of them, so the implication does not stand.
If $E$ has finite measure, then this can't happen. Suppose in fact that you have your sequence $E_i$ whose measures do not converge to $0$, so there is a subsequence whose measure converges to $$L = \limsup \mu(E_i) > 0.$$
On the other side, the set $E^n$ of the points that haven't disappeared at stage $n$ is defined by
$$
E^n = \bigcup_{m>n} E_m
$$
so it's a decreasing sequence of sets. But their measure is bounded (above and below):
$$
0 < L < \mu(E^n) ≤ \mu(E) < \infty
$$
so the limit set
$$
E^\infty = \bigcap_{n>0} \bigcup_{m>n} E_m
$$
has positive measure and so it is not empty.
A: The conjecture is true for $m(E) < \infty$. This is a rather general fact from measure theory. A proof goes as follows.
Define
$$
F_n := \bigcup_{j=n}^\infty E_j\quad for\ n\in\mathbb N.
$$
By construction we have $F_n = E_n \cup F_{n+1}$ and thus $E_n \subseteq F_n$ and $m(E_n) \leq m(F_n)$. Moreover, we see that the $F_n$ are a descending sequence, i.e. $\forall n\in \mathbb N:F_{n+1} \subseteq F_n.$ And finally, the assumption on the $E_n$ implies
$$
\forall x\in E:\exists N_x \in \mathbb N:\forall k > N_x:x \notin F_k.
$$
This means that
$$
\bigcap_{k=1}^\infty F_k = \emptyset.
$$
Now we can use the continuity from above of measures (see here: http://en.wikipedia.org/wiki/Measure_(mathematics)#Measures_of_infinite_intersections_of_measurable_sets). We can do this because $m(E)$ is finite. The continuity from above follows easily from the $\sigma$-additivity of measures. The continuity from above gives us
$$
0 = m(\emptyset) = m(\bigcap_{k=1}^\infty F_k) = \lim_{n\rightarrow\infty}m(F_n).$$
This together with the fact $0\leq m(E_n)\leq m(F_n)$ from above implies
$$
0 = \lim_{n\rightarrow\infty} E_n,
$$
as desired.
