How do I prove that
$$1_{A∪B} = 1_A + 1_B – 1 _{A∩B}$$ ?
For the proof of intersection I found on mathexchange that:
\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}
I tried to do the same thing as in writing $ 1_A + 1_B – 1 _{A∩B}$
out like that but ended up getting really confused and can't seem to be able to prove this.
\begin{align}1_A(x)+1_B(x)-1_A(x)1_B(x)&=\begin{cases}…+\begin{cases}…-\begin{cases}
\end{cases}\end{cases}\end{cases}…\end{align}
Best Answer
Distinguish between the following cases:
These are all disjoint cases and $A\cup B= (A\setminus B) \cup (B\setminus A) \cup (A\cap B)$. Now, if $1_{A\cup B}(x)=1$ then only one of the above holds. You can then verify that both sides are equal in $1_{A\cup B}= 1_A+1_B-1_{A\cap B}$ for each of these three cases and you are done since then the functions have equal values for all $x$ (trivially if the left hand side is zero then the right hand side is also).