On pointwise convergence of Lebesgue measurable sets and some properties Definition (Pointwise convergence of sets): We say that a sequence $E_n$ of sets in $\mathbb{R}^d$ converges pointwise to another set $E$ in $\mathbb{R}^d$ if the indicator functions $1_{E_n}$ converge pointwise to the indicator function $1_E$.

The problem is to show that if $E_n$ are all Lebesgue measurable, and converge pointwise to $E$, then $E$ is also Lebesgue measurable.
Further, I need to show that $m(E_n)$ converges to $m(E)$ [where $m(\cdot)$ denotes the Lebesgue measure], if $E_n$ are all contained in another Lebesgue measurable set $F$ of finite measure. Also, I've to produce a counter example to show that the convergence may not hold without the assumption.

For the first part, I tried to write $E$ as some countable union/intersection of the $E_n$'s. But I cannot exploit the definition of pointwise convergence of sets well enough to defend my point. For the second part, probably I need to incorporate the monotone convergence theorem, but I don't have a clear-cut attack on the problem. Any help would be greatly appreciated!
 A: A standard result is that the pointwise limit of a sequence of measurable functions is itself measurable, provided the ambiant space is complete (and $\mathbb{R}^d$ is complete). Since
$$1_{E_n}(x)\rightarrow 1_{E}(x)$$
for every $x$ and $1_{E_n}$ is measurable, it follows that $1_E$ is also measurable, hence $E$ is a measurable set.
For the second part, we use the dominated convergence theorem. We know that
$$1_{E_n}(x)\rightarrow 1_{E}(x)$$
for every $x$, while
$$|1_{E_n}(x)|\leq 1_F(x)$$
which is an integrable function because $F$ has finite measure. It follows that
$$m(E_n)=\int 1_{E_n}(x)dx\rightarrow\int 1_E(x)dx=m(E)$$
as $n\rightarrow\infty$.
A: This question is quite similar to Exercise 1.2.13 from Tao's Measure Theory book.  I would like provide a more basic proof.
The first one is answered in Sequence of Measurable Sets Converge Pointwise to a Measurable Set. 
Let 
$$F_n:= \cap_{m\ge n} E_m, \text{ and } G_n:=\cup_{m\ge n} E_m.$$
We have $E = \cup_{n \ge 1} F_n$, because $1_E(x) = \liminf_{n\to\infty} 1_{E_n}(x) = \lim_{n\to \infty}1_{F_n}(x)$ and $F_1\subset F_2\subset \ldots \subset \mathbb{R}^d$.  Similarly $E = \cap_{n\ge 1} G_n$.
With these observations, we can prove the first part.  Because each $F_n$ is a countable intersection of measurable sets $(E_n)$, each $F_n$ is measurable.  Similarly, because $E$ is a countable union of measurable sets $(F_n)$, $E$ is measurable.  (Lemma 1.2.13)
For the second part, because $F_n$ is a family of countable non-decreasing sequence of measurable sets, we can apply upward monotone convergence for measurable sets (Exercise 1.2.11), and have 
$$\lim_n \mu(F_n) = \mu(\cup_n F_n) = \mu(E)$$
where $\mu$ denote the Lebesgue measure.  Additionally, for all $n\ge 1$, because $F_n\subset E_n$, $\mu(F_n)\le \mu(E_n)$, and $\mu(E)\le \liminf_n\mu(E_n)$.
Similarly, Because $(G_n)$ is a countable non-increasing sequence of measurable sets, and $\mu(G_1)$ is finite because all $E_n$ are contained in another measurable set of finite measure, we can apply downward monotone convergence for measurable set.  Thus 
$$\lim_{n\to \infty} \mu(G_n) = \mu(\cap_{n} G_n) = \mu(E).$$
Additionally, for all $n\ge 1$, because $G_n\supset E_n$, $\mu(F_n)\ge \mu(E_n)$, and $\mu(E)\ge \limsup_n\mu(E_n)$.  This completes the proof.
Finally, consider $E_n = [n, n+1]$ for all $n\ge 1$ and $E = \emptyset$.
