# 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 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 $n\rightarrow\infty$.