Proof of the union of two indicator functions 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}
 A: Distinguish between the following cases:


*

*$x\in A\setminus B$,

*$x\in B\setminus A$,

*$x\in A\cap B$.


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).
A: It's simple: 


*

*on the complementary set of $A\cup B$, $1_A, 1_B$ and $1_{A\cap B}$ are all $0$;

*on  $A\setminus B$ only $1_A$ is one. Similarly, on $B\setminus A$, only $1_B$ is $1$;

*on $A\cap B$,  $1_A, 1_B$ and $1_{A\cap B}$ are all $1$.

A: Use that if you have disjoint sets, such as $A$ and $A^\complement\cap B$, then the indicator for the union equals the sum of the indicators for the disjoint sets. $$\mathbf 1_{A\cup (A^\complement\cap B)}=\mathbf 1_{A}+\mathbf 1_{A^\complement\cap B}$$
Because an element cannot be in both sets, by definition of disjoint, so if $\mathbf 1_A=1$ then $\mathbf 1_{A^\complement\cap B}=0$ and vice versa, therefore their sum is $\mathbf 1_{A\cup(A^\complement\cap B)}$ 

Now simply note that $A\cup B$ equals $(A\cap B^\complement)\cup(A\cap B)\cup(A^\complement \cap B)$ which is a union of disjoint sets.
