$\{A \subset X: \chi_A \in \mathcal{F}\}$ is a sigma algebra Suppose $\mathcal{F}$ is a collection of real-valued functions on $X$ such that the constant functions are in $\mathcal{F}$ and $f + g$, $fg$, and $cf$ are in $\mathcal{F}$ whenever $f$, $g \in \mathcal{F}$ and $c \in \mathbb{R}$. Suppose $f \in \mathcal{F}$ whenever $f_n \to f$ and each $f_n \in \mathcal{F}$. Define the function$$\chi_A(x) = \begin{cases} 1 & x \in A \\ 0 & x \notin A.\end{cases}$$Does it follow that $\mathcal{A} = \{A \subset X : \chi_A \in \mathcal{F}\}$ is a $\sigma$-algebra?
 A: $\emptyset \in \cal A$ because $\chi_\emptyset \equiv 0$.
If $A \in \cal A$ then $X \setminus A \in \cal A$  because $\chi_{X \setminus A} = 1 - \chi_A = 1 + (-1) \cdot \chi_A$.
If $A,B \in \cal A$ then $A \cap B \in \cal A$ because $\chi_{A \cap B} = \chi_A \chi_B$.
DeMorgan's rules and that fact that $\cal A$ is closed under complementation imply that $A \cup B \in \cal A$ whenever $A,B \in \cal A$. A simple induction argument leads to the conclusion that $A_1\cup \cdots\cup A_n \in \cal A$ whenever $A_1,\ldots,A_n \in \cal A$.
Finally observe that $$\chi_{\cup_{k=1}^\infty A_k} = \lim_{n \to \infty} \chi_{\cup_{k=1}^n A_k}$$
because the sequence $\chi_{\cup_{k=1}^n A_k}$ is nondecreasing and $0$-$1$ valued. If $\{A_k\}$ is a sequence in $\cal A$ it follows that $\cup A_k \in \cal A$ too.
A: Yes, $\mathcal{A}$ is a $\sigma$-algebra.


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*It is clear that $\chi_{X} =1 \in \mathcal{F}$ since it is a constant function, so $X\in \mathcal{A}$

*Note that $\chi_{A\cap B} = \chi_{A} \chi_{B}$ so if $A,B\in \mathcal{A}$ it follows that $A\cap B\in \mathcal{A}$. Then $\chi_{A\cup B} = \chi_{A} +\chi_{B} - \chi_{A\cap B} \in \mathcal{F}$. Also $\chi_{A\setminus B}=\chi_{A}-\chi_{A\cap B} \in \mathcal{F}$ so $A\cup B$,$A\setminus B \in \mathcal{A}$

*If $E_k$ are disjoint, then $f_n:=\chi_{\bigcup_{k=1}^{n} E_k} = \sum_{k=1}^n \chi_{E_k} \in \mathcal{F}$ and $f_n\to f:=\chi_{\bigcup_{k=1}^{\infty}E_k}\in \mathcal{F}$ so $\bigcup_{k} E_k \in \mathcal{A}$

