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A book I'm reading says the following about indicator functions $\chi_A$ :

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But unless I'm missing something, how can that that be? If $B$ is for example the set $(-2,2)$, $1 \in B$, but since $B$ contains both $0$ and $1$, $\chi_A^{-1}(B) = \{x \in A\} \cup \{x \not \in A\} = S \neq A$. Similarly, if $B$ is, say, the set $(-1, 0.5)$, then even though $1 \not \in B$, $\chi_A^{-1}(B)= \{x \not \in A\} \neq \{\emptyset\}$.

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up vote 2 down vote accepted

You are correct. The statement should be

$$\chi_A^{-1}(B) = \begin{cases} \emptyset & 0, 1 \not\in B\\ A & 0 \not\in B, 1 \in B\\ S\setminus A & 0 \in B, 1 \not\in B\\ S & 0, 1 \in B. \end{cases}$$

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