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Here is the problem: Prove that $I_{A_1\cup A_2 \cup\dots\cup A_n}(\omega)=\text{max}\{I_{A_2}(\omega),I_{A_1}(\omega),\dots,I_{A_n}(\omega)\}$ for any sets $A_1,A_2,\dots,A_n$

Here is what I know: $$I_{A_1\cup A_2 \cup\dots\cup A_n}(\omega)\begin{gathered}=\begin{cases} 1\quad\text{if}\quad\omega\in A_1\cup A_2 \cup\dots\cup A_n \\ 0\quad\text{if}\quad\omega\notin A_1\cup A_2 \cup\dots\cup A_n \end{cases}\\=\begin{cases} 1\quad\text{if}\quad\omega~\text{belongs in one of the}~A_n's \\ 0\quad\text{if}\quad\omega~\text{belongs in none of the}~A_n's\end{cases}\\=\begin{cases} 1\quad\text{if}\quad I_{A_n}(\omega)=1~\text{for at least one}~A_n \\ 0\quad\text{if}\quad I_{A_n}(\omega)=0~\text{for all}~A_n's\end{cases}\end{gathered}$$

I don't know where to go from here.

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  • $\begingroup$ You can proceed by formulating a similar casewise definition of the max function on the RHS. Tip: write $A_i$ instead of $A_n$ since $n$ denotes the number of sets. $\endgroup$
    – Shubham Johri
    Commented Mar 10, 2021 at 5:58
  • $\begingroup$ It's fairly clear that your last expression is the desired maximum, so I'd just add $=\max\{...\}$ as a final line. $\endgroup$
    – Karl
    Commented Mar 10, 2021 at 5:58

1 Answer 1

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You can proceed by formulating a similar casewise definition of the max function on the RHS, i.e.$$\max_i\{I_{A_i}(\omega)\}=\begin{cases}1,&\exists A_j|I_{A_j}(\omega)=1\\0,&I_{A_k}(\omega)=0\forall k\end{cases}$$which is identical to the function definition you wrote.

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