The $L^p$ limit of characteristic functions is a characteristic function.

Question

Let $(X,\mathcal{A},\mu)$ a measurable space where $X$ is a set, $\mathcal{A}$ is a $\sigma$-álgebra and $\mu:\mathcal{A}\rightarrow[0,\infty]$ is a measure. Let $(\chi_{E_{n}})$ be a sequence of characteristic functions; this is
$$\chi_{E_{n}}(x)=\begin{cases}
0,&x\notin E_n,\\
1, &x\in E_n.
\end{cases}$$
where $E_n\in \mathcal{A}$. Suppose that the sequence $(\chi_{E_{n}})$ is a Cauchy sequence in $L^p(X,\mathcal{A},\mu)$; that is given $\epsilon>0$ exists $n_0\in \mathbb{N}$ such that
$$\|\chi_{E_n}-\chi_{E_{n}} \|_p=\left(\int|\chi_{E_{n}}-\chi_{E_{m}}|^pd\mu\right)^{1/p}<\epsilon,\,\forall n\geq n_0.$$
I would like to probe that exists a function $f\in L^p$ such that $f=\chi_{E}$ where $E\in \mathcal{A}$ such that the sequence $(\chi_{E_{n}})$ converges to $f$ in $L_p$; that is, given $\varepsilon>0$ exists $n_1>0$ such that
$$\|\chi_{E_{n}}-f\|_p<\varepsilon,\,\forall n\geq n_1.$$
My attempt: I think the function is $f=\chi_{E}$ where
$$E=\bigcup_{m=1}^{\infty}\left(\bigcap_{k=m}^{\infty}E_k\right).$$
The reason I believe it is that the identity
$$|\chi_{E_{n}}-\chi_{E_{m}}|=\chi_{E_{n}\triangle E_{m}}$$
where $E_{n}\triangle E_{m}=E_n\backslash E_m\cup E_m\backslash E_n$ together with the Cauchy property implies that
$$\mu(E_{n}\triangle E_{m})<\epsilon$$
So my reasoning tells me that the "substance" of the functions is in their intersection.

Answer 1

Let $f$ be the limit of $\chi_{E_n}$ in $L^p$. All you have to do is show that $f(x) = 0$ or $f(x) = 1$ almost everywhere. Then we see that $f = \chi_E$ almost everywhere, where $E = \{x \in X | f(x) = 1\}$

To do this, it is helpful to define $g(x) = \min(|x - 1|, |x|)$. We will show that $g \circ f = 0$ almost everywhere.

We first show that $\int g(f(x))^p \; dx = 0$. Since $g(f(x))^p \geq 0$, it suffices to show that the integral is not positive. Indeed, consider an arbitrary $\epsilon > 0$, and takesome $n$ such that $\int |\chi_{E_n} - f|^p dx < \epsilon$. Then we note that $|\chi_{E_n}(x) - f(x)| \geq \min(|f(x) - 1|, |f(x)|) = g(f(x))$. Then we see that $\int g(f(x))^p \; dx \leq \int |\chi_{E_n}(x) - f(x)|^p dx < \epsilon$. This shows $\int g(f(x))^p \; dx = 0$.

Since $g(f(x))^p \geq 0$ for all $x$, this means that $g(f(x))^p$ is 0 almost everywhere; thus, $g \circ f$ is zero almost everywhere.

Since $g \circ f = 0$ almost everywhere, we see that $f(x) = \chi_{E}$ almost everywhere, where $E$ is defined as $\{x \in X| f(x) = 1\}$.