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 $\{x \in X| f(x) = 1\}$.

Answer 2

Completeness of $L^{p}$ tells you that thre exists $f \in L^{p}$ with $f_n \to f$ in $L^{p}$. This implies that there is a subsequence $(f_{n_k})$ of $(f_n)$ converging a.e. to $f$. A sequence of $0$'s and $1$'s can only converge to $0$ or $1$. Hence $f =0$ or $1$ a.e. Let $E=\{x: f(x)=1\}$ and show that $f=\chi_E$ a.e.

Answer 3

This problem is identical to Bartle's 7I in his elements of integration and Lebesgue measure. I state a couple of facts which are centainly in Bartle's book.

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Lemma 1: $L_p$ is complete.

Lemma 2: $L_p$ convergence implies convergence in measure.

This is straightforward: $$0\leftarrow\int |f-f_n|^p d\mu\geq \alpha^p \mu(\{x\::\: |f-f_n(x)|\geq \alpha\})\Rightarrow \lim_n \mu(\{x\: :\:|f-f_n(x)|\geq \alpha\})=0$$

Lemma 3: If a sequence converges in measure $f_n \xrightarrow[]{\mathcal{M}}f$, there is a subsequence $f_{n_k}$ which converges pointwise to $g$ almost everwhere. $f$ and $g$ agree almost everywhere.

This follows from Theorem 7.6 and its corollary from Bartle. Indeed, if $f_n$ converge in measure, in particular they are Cauchy in measure and hence $f_{n_k}\rightarrow g$ pointwise by Theorem 7.6. In the corollary Bartle proves $g$ and $f$ coincide almost everywhere.

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With these three facts it is easy to finish things off. By lemma $1$, there exists an $f\in L_p$ such that $f_n\xrightarrow[]{L_p} f$. By Lemma 2, $f_n\xrightarrow[]{\mathcal{M}} f$. Finally, by Lemma 3, we have that $f_{n_k}(x)\rightarrow g(x)$ pointwise almost everwhere. It happens that $f_{n_k}(x)$ is either $1$ or $0$, so for it to converge, we must have $\lim f_{n_k}=g=\chi_E$. $\chi_E$ is measurable being the limit of a sequence of measurable functions, thus $E$ is in the sigma algebra. Thus, $f$ and $\lim f_{n_k}=g=\chi_E$ coincide almost everywhere.