Let $\mathfrak{C}$ be a collection of subsets of a set $X$. Let $\mu$ be a nonnegative extended real-valued set function on $\mathfrak{C}$.
We say that $\mu$ is finitely additive on $\mathfrak{C}$ if $\mu(\cup_{k=1}^nE_k)=\sum_{k=1}^n\mu(E_k)$ for every disjoint finite sequence $(E_k:k=1,…,n)$ in $\mathfrak{C}$ such that $\cup_{k=1}^nE_k\in \mathfrak{C}$.
We say that $\mu$ is countably additive on $\mathfrak{C}$ if $\mu(\cup_{k=1}^{\infty}E_k)=\sum_{k=1}^{\infty}\mu(E_k)$ for every disjoint sequence $(E_k:k=1,2,3,…)$ in $\mathfrak{C}$ such that $\cup_{k=1}^{\infty}E_k\in \mathfrak{C}$.
Question 1: Does countable additivity imply finite additivity?
My thoughts: From one textbook: Assume that 1) $\emptyset\in \mathfrak{C}$ and 2) $\mu(\emptyset)=0$. Then countable additivity of $\mu$ implies its finite additivity. I don't understand why we need to assume this. The empty set is always the subset of any set by definition. Also we can prove that empty set possesses every property. So why we must have these assumptions?
Best Answer
Notice that $\mathfrak{C}$ is just any collection of subsets of $X$. You could, for example, have that $\mathfrak{C}=\{X\}$, or $\mathfrak{C}=\mathcal{P}(X)$, or $\mathfrak{C}=\{\varnothing\}$. For this reason, you have no reason to expect that $\varnothing\in\mathfrak{C}$. Indeed this motivates the definition of a $\sigma$-algebra, which we can define as a collection $\mathfrak{C}$ of subsets of $X$ such that $\varnothing\in\mathfrak{C}$ (or equivalently $X\in\mathfrak{C}$, or simply that $\mathfrak{C}$ is nonempty), and $\mathfrak{C}$ is closed under complementation and countable union (equivalently countable intersection).
Now that we have the definition of a $\sigma$-algebra, we can define a measure on a $\sigma$-algebra. So consider a set $X$ and a $\sigma$-algebra $\mathfrak{C}$ on $X$. We define a measure as a function $\mu:\mathfrak{C}\to[0,\infty]$ which is countably additive and such that $\mu(\varnothing)=0$. From this definition, one can deduce what you are asking about: that $\mu$ is also finitely additive. Indeed, let $A_1,\dots,A_n\in\mathfrak{C}$ be pairwise disjoint, and construct the sequence $\{\Omega_j\}_{j\in\mathbb{Z}^+}$ such that $\Omega_j=A_j$ for $j\in\{1,\dots,n\}$, and $\Omega_j=\varnothing$ otherwise. Then countable additivity gives us that
$$\mu\left(\bigcup_{j=1}^n A_j\right)=\mu\left(\bigcup_{j\in\mathbb{Z}^+} \Omega_j\right)=\sum_{j=1}^\infty\mu(\Omega_j)=\sum_{j=1}^n \mu(A_j),$$
which is finite additivity. But now notice how this proof required that $\mu(\varnothing)=0$ (hence why we include that in our definition of a measure), and for this to even make sense, that $\varnothing$ is in our $\sigma$-algebra. Indeed for our unions to even make sense to begin with, we needed our $\sigma$-algebra to be closed under countable union. All this motivates why we define there things the way we do; because they give rise to the properties we want and expect. Hopefully this answers your question.