Proof explanation: $I_A$ is measurable$ \iff A$ is measurable

Question

We look at $I_A^{-1}(-\infty,\alpha)$

$$I_A^{-1}(-\infty,\alpha) = \begin{cases} \mathbb{R}, & \alpha\geq1 \\ \emptyset, & \alpha<0 \\ A^C, & \text{for } 0 < \alpha < 1 \end{cases}$$

So $A^C$ is measurable $\iff$ $A$ is measurable

The optional images are $\{0,1\}$ so if the image is $(-\infty,0)$ then the pre image is an empty set Why the preimage when $0<\alpha<1$ is $A^C$ and $\mathbb{R}$ for $\alpha\geq1$?

Answer 1

The inverse image of $(-\infty,\alpha)$ is the set of all points in $\mathbb{R}$ at which $I_A$ takes values in $(-\infty,\alpha)$ Since $I_A$ only takes the values $0$ and $1,$ if $\alpha\ge1$ this is true of every point in $\mathbb{R}.$ If $0\le\alpha<1$ the only possible value of $I_a$ in $(-\infty,\alpha)$ is $0$ which is attained precisely at the points in $A^C$.

Answer 2

• $x \in A \iff I_A(x) = 1$, so $I_A^{-1}(\{1\}) = A$
• $x \notin A \iff I_A(x) = 0$, so $I_A^{-1}(\{0\}) = A^C$

 

1. When $\alpha \in (0,1)$, $I_A^{-1}((-\infty,\alpha)) \supseteq I_A^{-1}(\{0\}) = A^C$. To show the reverse set inclusion, let $I_A(x) < \alpha < 1$, so $x \notin A$ from above.
2. What $\alpha \ge 1$, $I_A^{-1}((-\infty,\alpha)) \supseteq I_A^{-1}(\{0,1\}) = \Bbb{R}$, so $I_A^{-1}((-\infty,\alpha) = \Bbb{R}$.