[proofplan]
We prove the four identities pointwise. For an arbitrary $\omega \in \Omega$, each indicator takes only the values $0$ and $1$, so each identity reduces to checking the membership status of $\omega$ in $A$ and $B$. The complement identity uses a two-case split, while the intersection, union, and set-difference identities use the four possible membership cases for the pair $(A,B)$.
[/proofplan]
[step:Declare the pointwise indicator convention]
For any event $C \in \mathcal F$, let
\begin{align*}
\mathbb{1}_C:\Omega \to \mathbb R
\end{align*}
denote the indicator function of $C$, defined by $\mathbb{1}_C(\omega)=1$ when $\omega \in C$ and $\mathbb{1}_C(\omega)=0$ when $\omega \notin C$. Since $A,B \in \mathcal F$, the sets $A^c$, $A\cap B$, $A\cup B$, and $A\setminus B$ also belong to $\mathcal F$, so all indicators appearing below are real-valued functions on $\Omega$.
[/step]
[step:Prove the complement identity by evaluating at an arbitrary point]
Fix $\omega \in \Omega$. If $\omega \in A$, then $\omega \notin A^c$, so
\begin{align*}
\mathbb{1}_{A^c}(\omega)=0=1-1=1-\mathbb{1}_A(\omega).
\end{align*}
If $\omega \notin A$, then $\omega \in A^c$, so
\begin{align*}
\mathbb{1}_{A^c}(\omega)=1=1-0=1-\mathbb{1}_A(\omega).
\end{align*}
Thus $\mathbb{1}_{A^c}(\omega)=1-\mathbb{1}_A(\omega)$ for every $\omega \in \Omega$, hence $\mathbb{1}_{A^c}=1-\mathbb{1}_A$ as functions on $\Omega$.
[guided]
We prove equality of functions by evaluating both sides at the same arbitrary input. Fix $\omega \in \Omega$. There are exactly two possibilities for membership in $A$.
If $\omega \in A$, then by the definition of complement, $\omega \notin A^c$. Therefore the indicator values are $\mathbb{1}_A(\omega)=1$ and $\mathbb{1}_{A^c}(\omega)=0$. Substituting these values into the proposed identity gives
\begin{align*}
\mathbb{1}_{A^c}(\omega)=0=1-1=1-\mathbb{1}_A(\omega).
\end{align*}
If $\omega \notin A$, then by the definition of complement, $\omega \in A^c$. Hence $\mathbb{1}_A(\omega)=0$ and $\mathbb{1}_{A^c}(\omega)=1$, so
\begin{align*}
\mathbb{1}_{A^c}(\omega)=1=1-0=1-\mathbb{1}_A(\omega).
\end{align*}
The point $\omega$ was arbitrary. Therefore the two functions $\mathbb{1}_{A^c}$ and $1-\mathbb{1}_A$ agree at every point of $\Omega$, which is precisely the statement that $\mathbb{1}_{A^c}=1-\mathbb{1}_A$ pointwise on $\Omega$.
[/guided]
[/step]
[step:Prove the intersection identity by checking joint membership in $A$ and $B$]
Fix $\omega \in \Omega$. If $\omega \in A\cap B$, then $\omega \in A$ and $\omega \in B$, so
\begin{align*}
\mathbb{1}_{A\cap B}(\omega)=1=1\cdot 1=\mathbb{1}_A(\omega)\mathbb{1}_B(\omega).
\end{align*}
If $\omega \notin A\cap B$, then at least one of $\omega \notin A$ or $\omega \notin B$ holds. Hence at least one of $\mathbb{1}_A(\omega)$ and $\mathbb{1}_B(\omega)$ is $0$, and therefore
\begin{align*}
\mathbb{1}_{A\cap B}(\omega)=0=\mathbb{1}_A(\omega)\mathbb{1}_B(\omega).
\end{align*}
Thus $\mathbb{1}_{A\cap B}=\mathbb{1}_A\mathbb{1}_B$ pointwise on $\Omega$.
[/step]
[step:Prove the union identity by exhausting the four membership cases]
Fix $\omega \in \Omega$. If $\omega \in A$ and $\omega \in B$, then $\omega \in A\cup B$ and $\omega \in A\cap B$, so
\begin{align*}
\mathbb{1}_{A\cup B}(\omega)=1=1+1-1=\mathbb{1}_A(\omega)+\mathbb{1}_B(\omega)-\mathbb{1}_{A\cap B}(\omega).
\end{align*}
If $\omega \in A$ and $\omega \notin B$, then $\omega \in A\cup B$ and $\omega \notin A\cap B$, so
\begin{align*}
\mathbb{1}_{A\cup B}(\omega)=1=1+0-0=\mathbb{1}_A(\omega)+\mathbb{1}_B(\omega)-\mathbb{1}_{A\cap B}(\omega).
\end{align*}
If $\omega \notin A$ and $\omega \in B$, then $\omega \in A\cup B$ and $\omega \notin A\cap B$, so
\begin{align*}
\mathbb{1}_{A\cup B}(\omega)=1=0+1-0=\mathbb{1}_A(\omega)+\mathbb{1}_B(\omega)-\mathbb{1}_{A\cap B}(\omega).
\end{align*}
If $\omega \notin A$ and $\omega \notin B$, then $\omega \notin A\cup B$ and $\omega \notin A\cap B$, so
\begin{align*}
\mathbb{1}_{A\cup B}(\omega)=0=0+0-0=\mathbb{1}_A(\omega)+\mathbb{1}_B(\omega)-\mathbb{1}_{A\cap B}(\omega).
\end{align*}
Therefore $\mathbb{1}_{A\cup B}=\mathbb{1}_A+\mathbb{1}_B-\mathbb{1}_{A\cap B}$ pointwise on $\Omega$.
[/step]
[step:Prove the set-difference identity by checking membership in $A$ and exclusion from $B$]
Fix $\omega \in \Omega$. If $\omega \in A\setminus B$, then $\omega \in A$ and $\omega \notin B$, so
\begin{align*}
\mathbb{1}_{A\setminus B}(\omega)=1=1(1-0)=\mathbb{1}_A(\omega)(1-\mathbb{1}_B(\omega)).
\end{align*}
If $\omega \notin A\setminus B$, then either $\omega \notin A$ or $\omega \in B$. In the first case $\mathbb{1}_A(\omega)=0$, and in the second case $1-\mathbb{1}_B(\omega)=0$. Hence the product $\mathbb{1}_A(\omega)(1-\mathbb{1}_B(\omega))$ is $0$ in either case, and
\begin{align*}
\mathbb{1}_{A\setminus B}(\omega)=0=\mathbb{1}_A(\omega)(1-\mathbb{1}_B(\omega)).
\end{align*}
Thus $\mathbb{1}_{A\setminus B}=\mathbb{1}_A(1-\mathbb{1}_B)$ pointwise on $\Omega$. Combining the four pointwise identities proves the theorem.
[/step]
