[proofplan] The complement $A^c$ is measurable because $\mathcal F$ is a $\sigma$-algebra. The sets $A$ and $A^c$ are disjoint and their union is the whole sample space $\Omega$. Applying finite additivity of the probability measure to this disjoint union gives $\mathbb P(\Omega)=\mathbb P(A)+\mathbb P(A^c)$, and the normalization axiom $\mathbb P(\Omega)=1$ gives the desired identity. [/proofplan] [step:Verify that the complement is a measurable event] Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space from the statement, so $\Omega$ is the sample space, $\mathcal F$ is a $\sigma$-algebra on $\Omega$, and $\mathbb P:\mathcal F\to[0,1]$ is a probability measure. Let $A\in\mathcal F$ be the event from the statement, and define the complement of $A$ in $\Omega$ by $A^c:=\Omega\setminus A$. Since $\mathcal F$ is a $\sigma$-algebra and $A\in\mathcal F$, closure under complements in the [definition of a sigma-algebra](/page/Sigma-Algebra) gives $A^c\in\mathcal F$. [/step] [step:Decompose the sample space as a disjoint union] Let $\varnothing$ denote the empty subset of $\Omega$. By the definition of complement, \begin{align*} A\cap A^c=\varnothing \end{align*} and \begin{align*} A\cup A^c=\Omega. \end{align*} Thus $A$ and $A^c$ are disjoint measurable subsets of $\Omega$ whose union is $\Omega$. [/step] [step:Apply finite additivity and probability normalization] Because $A\in\mathcal F$, $A^c\in\mathcal F$, and $A\cap A^c=\varnothing$, finite additivity of the probability measure $\mathbb P$ gives \begin{align*} \mathbb P(A\cup A^c)=\mathbb P(A)+\mathbb P(A^c). \end{align*} Using $A\cup A^c=\Omega$ and the normalization axiom for a probability measure, $\mathbb P(\Omega)=1$, we obtain \begin{align*} 1=\mathbb P(\Omega)=\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*} [guided] We now use the two defining facts about a probability measure that are relevant here: additivity on disjoint measurable sets and normalization on the whole sample space. We have already verified that $A\in\mathcal F$ and $A^c\in\mathcal F$, and we have shown that \begin{align*} A\cap A^c=\varnothing. \end{align*} Therefore the hypotheses for finite additivity are satisfied: the two sets are measurable and disjoint. Applying finite additivity of the probability measure $\mathbb P$ to the pair of sets $A$ and $A^c$ gives \begin{align*} \mathbb P(A\cup A^c)=\mathbb P(A)+\mathbb P(A^c). \end{align*} The point of using the pair $A$ and $A^c$ is that their union is exactly the entire sample space: \begin{align*} A\cup A^c=\Omega. \end{align*} Substituting this identity into the left-hand side gives \begin{align*} \mathbb P(\Omega)=\mathbb P(A)+\mathbb P(A^c). \end{align*} Since $\mathbb P:\mathcal F\to[0,1]$ is a probability measure, its normalization axiom says that the whole sample space has probability one: \begin{align*} \mathbb P(\Omega)=1. \end{align*} Combining the last two displayed identities yields \begin{align*} 1=\mathbb P(A)+\mathbb P(A^c). \end{align*} Finally, subtracting $\mathbb P(A)$ from both sides gives \begin{align*} \mathbb P(A^c)=1-\mathbb P(A). \end{align*} [/guided] [/step]