[proofplan]
We prove each identity directly from the probability axioms. First we derive $\mathbb P(\varnothing)=0$ and binary finite additivity from countable additivity by padding a finite disjoint family with empty sets. The complement formula follows from the disjoint decomposition $\Omega=A\cup A^c$, monotonicity follows from $B=A\cup(B\setminus A)$ when $A\subset B$, and inclusion-exclusion follows by comparing the two disjoint decompositions $A\cup B=A\cup(B\setminus A)$ and $B=(A\cap B)\cup(B\setminus A)$.
[/proofplan]
[step:Derive the empty-set identity and finite additivity from countable additivity]
Because $\mathcal F$ is a $\sigma$-algebra, $\varnothing\in\mathcal F$. Let $\mathbb N=\{1,2,3,\dots\}$ denote the set of positive integers. Apply countable additivity to the pairwise disjoint sequence $(E_n)_{n\in\mathbb N}$ defined by $E_1=\Omega$ and $E_n=\varnothing$ for every $n\ge 2$. Since $\bigcup_{n=1}^{\infty}E_n=\Omega$ and $\mathbb P(\Omega)=1$, we obtain
\begin{align*}
1=\mathbb P(\Omega)=\sum_{n=1}^{\infty}\mathbb P(E_n)=1+\sum_{n=2}^{\infty}\mathbb P(\varnothing).
\end{align*}
The terms in the final series are nonnegative, so $\sum_{n=2}^{\infty}\mathbb P(\varnothing)=0$, and therefore $\mathbb P(\varnothing)=0$.
Now let $E,F\in\mathcal F$ be disjoint events. Define a sequence $(G_n)_{n\in\mathbb N}$ of events by $G_1=E$, $G_2=F$, and $G_n=\varnothing$ for every $n\ge 3$. Countable additivity and $\mathbb P(\varnothing)=0$ give
\begin{align*}
\mathbb P(E\cup F)=\sum_{n=1}^{\infty}\mathbb P(G_n)=\mathbb P(E)+\mathbb P(F).
\end{align*}
Thus $\mathbb P$ is finitely additive on two disjoint events.
[guided]
The probability axioms give countable additivity, while the later arguments naturally use finite additivity. We therefore first derive the finite version carefully from the countable one.
Since $\mathcal F$ is a $\sigma$-algebra, it contains $\Omega$ and is closed under complements, so $\varnothing=\Omega^c$ belongs to $\mathcal F$. Let $\mathbb N=\{1,2,3,\dots\}$ denote the set of positive integers. Define a sequence of events $(E_n)_{n\in\mathbb N}$ by setting $E_1=\Omega$ and $E_n=\varnothing$ for every $n\ge 2$. These events are pairwise disjoint, because the only nonempty one is $E_1=\Omega$. Their union is $\Omega$. Countable additivity gives
\begin{align*}
\mathbb P(\Omega)=\sum_{n=1}^{\infty}\mathbb P(E_n).
\end{align*}
Using $\mathbb P(\Omega)=1$ and the definition of the sequence, this becomes
\begin{align*}
1=1+\sum_{n=2}^{\infty}\mathbb P(\varnothing).
\end{align*}
Each term $\mathbb P(\varnothing)$ is nonnegative by the probability axiom of nonnegativity. Hence the nonnegative series on the right must be $0$, so its first term is $0$. Therefore
\begin{align*}
\mathbb P(\varnothing)=0.
\end{align*}
We now derive the exact finite-additivity rule that will be used below. Let $E,F\in\mathcal F$ be disjoint events. Define $(G_n)_{n\in\mathbb N}$ by $G_1=E$, $G_2=F$, and $G_n=\varnothing$ for every $n\ge 3$. This is a pairwise disjoint sequence of events, and its union is $E\cup F$. Countable additivity gives
\begin{align*}
\mathbb P(E\cup F)=\sum_{n=1}^{\infty}\mathbb P(G_n).
\end{align*}
Substituting the definition of $G_n$ and using $\mathbb P(\varnothing)=0$, we obtain
\begin{align*}
\mathbb P(E\cup F)=\mathbb P(E)+\mathbb P(F).
\end{align*}
This is the two-set finite additivity formula for disjoint events.
[/guided]
[/step]
[step:Compute the probability of a complement from the disjoint decomposition of $\Omega$]
Let $A\in\mathcal F$. Since $\mathcal F$ is a $\sigma$-algebra, $A^c\in\mathcal F$. The events $A$ and $A^c$ are disjoint and satisfy $\Omega=A\cup A^c$. By finite additivity on disjoint events,
\begin{align*}
1=\mathbb P(\Omega)=\mathbb P(A\cup A^c)=\mathbb P(A)+\mathbb P(A^c).
\end{align*}
Subtracting $\mathbb P(A)$ from both sides gives
\begin{align*}
\mathbb P(A^c)=1-\mathbb P(A).
\end{align*}
[/step]
[step:Prove monotonicity by splitting the larger event into a disjoint union]
Assume $A,B\in\mathcal F$ and $A\subset B$. Since $\mathcal F$ is a $\sigma$-algebra, $B\setminus A=B\cap A^c\in\mathcal F$. The events $A$ and $B\setminus A$ are disjoint, and $B=A\cup(B\setminus A)$. Finite additivity gives
\begin{align*}
\mathbb P(B)=\mathbb P(A)+\mathbb P(B\setminus A).
\end{align*}
By nonnegativity, $\mathbb P(B\setminus A)\ge 0$, hence
\begin{align*}
\mathbb P(A)\le \mathbb P(B).
\end{align*}
[/step]
[step:Decompose $A\cup B$ and $B$ to prove inclusion-exclusion]
Let $A,B\in\mathcal F$. Since $\mathcal F$ is a $\sigma$-algebra, the events $A\cap B$ and $B\setminus A=B\cap A^c$ belong to $\mathcal F$. The events $A$ and $B\setminus A$ are disjoint, and
\begin{align*}
A\cup B=A\cup(B\setminus A).
\end{align*}
Finite additivity gives
\begin{align*}
\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B\setminus A).
\end{align*}
Also, the events $A\cap B$ and $B\setminus A$ are disjoint, and
\begin{align*}
B=(A\cap B)\cup(B\setminus A).
\end{align*}
Finite additivity gives
\begin{align*}
\mathbb P(B)=\mathbb P(A\cap B)+\mathbb P(B\setminus A).
\end{align*}
Therefore
\begin{align*}
\mathbb P(B\setminus A)=\mathbb P(B)-\mathbb P(A\cap B).
\end{align*}
Substituting this identity into the formula for $\mathbb P(A\cup B)$ yields
\begin{align*}
\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B).
\end{align*}
Together with the previous steps, this proves all four stated identities.
[/step]
