1. ## Help with theorem

I need help with understand a theorem.

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

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

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

2. ## Re: Help with theorem

You missed some cases, for example when both $0$ and $1$ are in $B$ (that's what the example $B=(-2,2)$ shows), and when neither $0$ nor $1$ is in $B$. You have to consider the four cases.

3. ## Re: Help with theorem

Case (i): $\{1\} \in B, \; \{0\} \notin B \; \Rightarrow 1_{A}^{-1}(B) = A$

Case (ii): $\{1\} \notin B, \; \{0\} \in B \; \Rightarrow 1_{A}^{-1}(B) = A^{c}$

Case (iii): $\{1\} \in B, \; \{0\} \in B \; \Rightarrow 1_{A}^{-1}(B) = E$

Case (iv): $\{1\} \notin B, \; \{0\} \notin B \; \Rightarrow 1_{A}^{-1}(B) = \emptyset$

Since $\mathcal{E}$ is a $\sigma$-algebra, all these sets are in $\mathcal{E}$. So $1_{A}$ is measurable.