Hey,

The indicator function is defined as:

$\displaystyle \mathbb{1}_{A}(x)=1$ if $\displaystyle x \in A$

$\displaystyle \mathbb{1}_{A}(x)=0$ if $\displaystyle x \notin A$

I want to show that: $\displaystyle \mathbb{1}_{A\cup B}(x)= \min (\mathbb{1}_{A}(x) + \mathbb{1}_{B}(x),1) $.

Here what I have done:

For every $\displaystyle x $ we have that $\displaystyle x \in A \cup B$ means that $\displaystyle x \in A$ or $\displaystyle x \in B$.

This means that $\displaystyle \mathbb{1}_{A}=1$ or $\displaystyle \mathbb{1}_{B}=1$.

How do I argue from here? I know the addition rule for mutual exclusive events. Maybe I am overthinking this?