Proving $\lim\limits_{N\to\infty}\mu\Bigl(\bigcup\limits_{n=N}^\infty(E\setminus E_n)\Bigr)=0$

Question

Let $E$ be a measurable set of $\mathbb{R}$, and the characteristic function of a subset $A$ of $E$ is a function$$\mathbf1_A:X\to\{0,1\}$$defined as$$\mathbf1_A(x):=\begin{cases}1&\text{ if }x\in A\\0&\text{ if }x\notin A\end{cases}.$$ Now let $\{E_{n}: n \in \mathbb{N}^+\} \subseteq E$ be a collection of subset of $E$. Show that
1. The sequence $\{\mathbf{1}_{E_n}(x)\}$ uniformly converges to $\mathbf{1}_{E}(x)$ if and only if $\exists N \in \mathbb{N}^+$, $\forall n \geq N$ such that $E=E_n$.

2. $\{\mathbf{1}_{E_n}(x)\}$ converges almost uniformly to $\mathbf{1}_{E}(x)$ if and only if $$\lim _{N \rightarrow \infty}\mu\left(\bigcup_{n=N}^\infty \left(E \backslash E_{n}\right)\right)=0,$$ where $\mu (⋅)$ denotes the Lebesgue measure.

3.The sequence $\{\mathbf{1}_{E_n}(x)\}$ converges to $\mathbf{1}_{E}(x)$ a.e. iff $$\mu\left(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty \left(E \backslash E_{n}\right)\right)=0.$$

My attempt:

1:$\Longleftarrow$ it is quite obvious.
$\Longrightarrow$ there exists a natural number $N$ such that $|\mathbf{1}_{E_n}(x)-\mathbf{1}_{E}(x)|<\epsilon$ for all $n \geq N$ and $x \in E$, namely $|\mathbf{1}_{E_n}(x)-\mathbf{1}_{E}(x)|=0$. Thus we obtain $E=E_n$ for all $n \geq N$ and $x \in E$.

2:$\Longrightarrow$ By definition of almost uniformly, we have for every $\delta>0$ there exists a measurable set $E_{\delta}$ with measure less than $\delta$ such that $\{E_{n}: n \in \mathbb{N}^+\} $ converges to $\mathbf{1}_{E}(x)$ uniformly on $E \backslash E_{\delta}$.

I'm completely stuck on how to proceed. Any help and comments will be appreciated.

Answer 1

Fix $\delta>0$ with $\delta<1$. As you say, there exists $E_\delta$ with $\mu(E_\delta)<\delta$ and $1_{E_n}\to 1_E$ uniformly on $E\setminus E_\delta$. Choose $N$ such that for all $n\geq N$, we have $|1_{E_n}-1_E|<1$; this forces $1_{E_n}=1_E$, on $E\setminus E_\delta$, for all $n\geq N$. Now \begin{align} \mu\left(\bigcup_{n\geq N}E\setminus E_n\right) &=\mu\left(E\setminus \bigcap_{n\geq N}E_n\right)\\[0.3cm] &=\mu\left((E\cap E_\delta)\setminus \bigcap_{n\geq N}E_n\right)+ \mu\left((E\setminus E_\delta)\setminus \bigcap_{n\geq N}E_n\right)\\[0.3cm] &\leq \mu(E_\delta)+0<\delta. \end{align} The measure of the second set is zero because, inside $E\setminus E_\delta$ and with $n\geq N$, each $E_n=E$. So the second term is $\mu((E\setminus E_\delta)\setminus E)=\mu(\varnothing)=0$.