[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.[/step]

[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: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]