Indicator functions, mean value, unclear property

Question

I have an easy question about indicator functions.

Let $A$ be any event. We can write $\Bbb P(A)$ as an expectation, as follows:

Define the indicator function:

$$ I_A = \begin{cases} 1, & \text{if event $A$ occurs} \\ 0, & \text{otherwise} \end{cases} $$

Then $I_A$ is a random variable. Why the first and last euality holds here:

$$ \Bbb E(I_A) = \sum_{r=0}^1 r \cdot \Bbb P(I_A = r) \\ = \Bbb P(A). $$

?

Answer 1

In general if $X$ is a random variable with $\mathsf P(X\in S)=1$ where $S$ denotes a countable set then: $$\mathsf EX=\sum_{s\in S}s\mathsf P(X=s)$$

Observe that $\mathsf P(1_A\in\{0,1\})=1$ and apply that to find the first equality.

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$\mathsf P(X=1)$ is just an abbreviation of $\mathsf P(\{\omega\in\Omega\mid X(\omega)=1\})$.

For $X=1_A$ we find: $$\{\omega\in\Omega\mid X(\omega)=1\}=\{\omega\in\Omega\mid 1_A(\omega)=1\}=A$$ so that: $$\mathsf P(1_A=1)=\mathsf P(A)$$

Answer 2

$r$ can take on two values, when $r=0$ the term vanishes, and when $r=1$ it takes on $P(I_A = 1) = P(A)$