# Help with theorem

• Aug 15th 2011, 04:49 AM
TheProphet
Help with theorem
I need help with understand a theorem.

Theorem Let $\displaystyle (F,\mathcal{F}) = (\mathbb{R},\mathcal{B}(\mathbb{R}))$, 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 $\displaystyle 1_{A} \neq 1$.
• Aug 15th 2011, 05:16 AM
girdav
Re: Help with theorem
You missed some cases, for example when both $\displaystyle 0$ and $\displaystyle 1$ are in $\displaystyle B$ (that's what the example $\displaystyle B=(-2,2)$ shows), and when neither $\displaystyle 0$ nor $\displaystyle 1$ is in $\displaystyle B$. You have to consider the four cases.
• Aug 16th 2011, 02:34 AM
TheProphet
Re: Help with theorem
Case (i): $\displaystyle \{1\} \in B, \; \{0\} \notin B \; \Rightarrow 1_{A}^{-1}(B) = A$

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

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

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

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