I have to prove that in an indicator function it is true that:
$$1_A\cup _B…\cup _n(x)=max[1_A(x),1_B(x),…,1_n(x)]$$
Can you help me?
I am able to prove to the intersection only, as below:
\begin{align}1_A(x)1_B(x)&=\begin{cases}
1& x\in A\\
0& x\in A^C
\end{cases}\begin{cases}
1& x\in B\\
0& x\in B^C
\end{cases}\\&=\begin{cases}
1& x\in A \cap x\in B\\
0\cdot 1& x\in A^C\cap B\\
1\cdot 0& x\in A \cap B^C\\
0\cdot 0& x\in A^C \cap B^C\\
\end{cases}\\&=\begin{cases}
1& x\in A \cap B\\
0& x\in \underbrace{(A^C\cap B)\cup(A \cap B^C )\cup(A^C \cap B^C)}_{=(A\cap B)^C}\\
\end{cases}\\&=1_{A\cap B}(x)\end{align}
thanks!
Best Answer
Suppose $A_i\subset \Omega$ for all $i$. We want to prove that $$ 1_{\bigcup_{i=1}^n A_i}(x)=\max_{i=1}^n[1_{A_i}(x)]\tag{1} $$ for all $x\in\Omega$. Consider the following cases. Given $x\in\Omega$, suppose $x\in A_i$ for some $i$. What is the value of the expressions on both sides of the equation (1)? Next, given $x\in\Omega$, suppose $x\notin A_i$ for all $i$. What is the value of the expressions on both sides of the equation (1)?