Definition The indicator of $A\subset X$ is a function $\chi_A:X\longrightarrow\{0,1\}$ defined as
$$
\chi_A(x):=\begin{cases}1\;\text{if}\;x\in A\\0\;\text{if}\;x\notin A\end{cases}
$$
Problem 1 Prove the following property for a characteristic function $\chi_A$ of a subset $A$ of a set $X$:
$$
|A|=\sum_{x\in X}\chi_A(x)
$$
Problem 2 Check if the following equation holds:
$$
\chi_{A\cup B}=\chi_A + \chi_B – \chi_A\chi_B
$$
I don't know how to deal with the first problem yet because I have no idea how to apply the definition above. Any hints on how to start are welcome. I'll edit this entry once I know what needs to be done here.
Edit 1 First I'll add Brian M. Scott's proof for Problem 1:
$$
\begin{align}
|A|&=\sum_{x\in A}\underbrace{\chi_A(x)}_{1}\\
&=\sum_{x\in A}\chi_A(x)+\sum_{x\in X\setminus A}\underbrace{\chi_A(x)}_{0}\\
&=\sum_{x\in (A\cup (X\setminus A))}\chi_A(x)\\
&=\sum_{x\in X}\chi_A(x).
\end{align}
$$
because $A\cup(X\setminus A)=(A\cup X)\setminus(A\setminus A)=X\setminus \emptyset=X$.
Applying our introduced Definition on Problem 2 multiple times results to
$$
\begin{align}
\chi_{A\cup B}(x)&=\begin{cases}1\;\text{if}\;x\in A\cup B\\0\;\text{if}\;x\notin A\cup B\end{cases}\\
&\overset{(*)}{=}\begin{cases}1\;\text{if}\;x\in A\lor x\in B\\0\;\text{if}\;x\notin A\land x\notin B\end{cases}\\
&=\begin{cases}1-0\;\text{if}\;x\in A\land x\notin B\\1-0\;\text{if}\;x\notin A\land x\in B\\1-0\;\text{if}\;x\in A\land x\in B\\1-1\;\text{if}\;x\notin A\land x\notin B\end{cases}\\
&=1-\begin{cases}0\;\text{if}\;x\in A\land x\notin B\\0\;\text{if}\;x\notin A\land x\in B\\0\;\text{if}\;x\in A\land x\in B\\1\;\text{if}\;x\notin A\land x\notin B\end{cases}\\
&=1-\left[\left(\begin{cases}1-0\;\text{if}\;x\notin A\\1-1\;\text{if}\;x\in A\end{cases}\right)\left(\begin{cases}1-0\;\text{if}\;x\notin B\\1-1\;\text{if}\;x\in B\end{cases}\right)\right]\\
&=1-\left[\left(1-\begin{cases}0\;\text{if}\;x\notin A\\1\;\text{if}\;x\in A\end{cases}\right)\left(1-\begin{cases}0\;\text{if}\;x\notin B\\1\;\text{if}\;x\in B\end{cases}\right)\right]\\
&=1-\big(1-\chi_A(x)\big)\big(1-\chi_B(x)\big)\\
&=\chi_A(x)+\chi_B(x)-\chi_A(x)\chi_B(x).
\end{align}
$$
by using $(*)$ as
$$
\begin{align}
\overline{A\cup B}&\Leftrightarrow \bigvee_{x\in X}\{x\notin (A\cup B)\}\\
&\overset{(*)}{\Leftrightarrow}\bigvee_{x\in X}\{x\notin A\land x\notin B\}\\
&\Leftrightarrow\overline{A}\cap\overline{B}.
\end{align}
$$
I've got the feeling there's an easier way to solve the second problem. I'd be glad if anyone could give me hints for a shorter solution for this one as well.
Edit 2 I came up with a new proof for the second problem. Mind that
$$
\begin{align}
\chi_{\overline{\overline{A}}}(x)&=1-\chi_{\overline{A}}(x)\\
&=1-(1-\chi_A(x))\\
&=\chi_A(x).
\end{align}
$$
which gave me the idea for considering
$$
\begin{align}
\chi_{\overline{\overline{A\cup B}}}(x)&=1-\chi_{\overline{A\cup B}}(x)\\
&\overset{(*)}{=}1-\chi_{\overline{A}\cap\overline{B}}(x)\\
&\overset{(\times)}{=}1-\chi_{\overline{A}}(x)\chi_{\overline{B}}(x)\\
&=1-\big(1-\chi_A(x)\big)\big(1-\chi_B(x)\big)\\
&=\chi_A(x)+\chi_B(x)-\chi_A(x)\chi_B(x).
\end{align}
$$
where I proved beforehand that $\chi_{A\cap B}(x)\overset{(\times)}{=}\chi_A(x)\chi_B(x)$ by applying the definition for characteristic functions multiple times.
Best Answer
For the first question you have simply
$$|A|=\sum_{x\in A}1=\sum_{x\in A}1+\sum_{x\in X\setminus A}0=\sum_{x\in X}\chi_A(x)\;,$$
assuming, of course, that $A$ is a finite set.
HINT: For the second question just compare the two sides for each $x\in X$. Note that each $x\in X$ is in exactly one of the sets $A\setminus B$, $B\setminus A$, $A\cap B$, and $X\setminus(A\cup B)$.