I need help with understand a theorem.

TheoremLet $\displaystyle (F,\mathcal{F}) = (\mathbb{R},\mathcal{B}(\mathbb{R}))$, and $\displaystyle (E,\mathcal{E})$ be any measurable space. An indicator $\displaystyle 1_{A} $ on $\displaystyle E $ is measurable if and only if $\displaystyle A \in \mathcal{E} $.

Proof: If $\displaystyle 1_{A} $ is measurable then $\displaystyle A = \{\omega \, : \, 1_{A}(\omega) = 1 \}$ is in $\displaystyle \mathcal{E} $. I guess we can see this by noticing that $\displaystyle \{1\} = \bigcap_{n=1}^{\infty} \left(1-\frac{1}{n},1+\frac{1}{n}\right) $, which is in $\displaystyle \mathcal{F} $ and thus also $\displaystyle 1_{A}^{-1}(\{1\}) \in \mathcal{E} $ by definition.

Now, conversely assume $\displaystyle A \in \mathcal{E} $. Then if $\displaystyle B \in \mathcal{F}, \, 1_{A}^{-1}(B) = A \, \mbox{ if } 1 \in B \mbox{ and } 1_{A}^{-1}(B) = A^{c} \mbox{ if } 1 \notin B $.

Clearly $\displaystyle A^{c} \in \mathcal{E} $ because $\displaystyle \mathcal{E} $ is a $\displaystyle \sigma $-algebra. But shouldn't it be $\displaystyle 1_{A}^{-1}(B) = A $ if $\displaystyle B = \{1\} $ ? By taking for example $\displaystyle B = (-2,2) $ you get a lot of points where $\displaystyle 1_{A} \neq 1 $.