Vtyjkyuk

I am self-studying from Richard Bass's Real Analysis book, available here.

I am not sure about Exercise 3.2:

Suppose $(X,mathcalA)$ is a measurable space and $mu$ is a
non-negative set function that is finitely additive and such that
$mu(varnothing) = 0$ and $mu(X)<infty$. Suppose that whenever $A_i$ is a sequence of sets in $mathcalA$ which decrease to $varnothing$, then $lim_itoinftymu(A_i)=0$. Show that $mu$ is a measure.

In order to show that $mu$ is a measure, all that remains to be shown is for any countable collection of pairwise disjoint sets $B_i_ i =1,2,3ldots$,
$$mu(cup_i B_i)=sum_i mu(B_i).$$

My attempt:

Let $B_i_ i =1,2,3ldots$ be a countable collection of pairwise disjoint sets. Then consider $cup_i=n^inftyB_i downarrow varnothing$. It follows that $lim_ntoinftymu(cup_i=n^inftyB_i)=0$. We can rewrite this as
$$lim_ntoinftymuleft(cup_i=1^inftyB_i-cup_i=1^n-1B_iright)=0,$$
so by the algebra of limits
$$muleft(cup_i=1^inftyB_iright)=lim_ntoinftymu(cup_i=1^n-1B_i)=lim_ntoinftymu(cup_i=1^n B_i)=lim_ntoinftysum_i=1^nmu(B_i)=sum_i=1^inftymu(B_i).$$

Questions:

(1) Is this proof correct? (In particular I am not sure that $cup_i=n^inftyB_i$ does decrease to $varnothing$.)

(2) I do not appear to have used $mu(X)<infty$ anywhere. Why is this needed?

You cannot write $mu (Asetminus B)$ as $mu (A)-mu (B)$ for infinite measures (for $B subset A$). That is where the assumption $mu (X) <infty $ is required. Otherwise your proof is correct. It should be clear that $cup_i=n^infty B_i$ is decreasing. Suppose there is a point $x$ in the intersection. Then $x$ would belong to infinitely many of the sets $B_i$. But no point can belong to more than one of these sets, so the intersection is empty.

As Kavi Rama Murthy already noted, you use $mu(X) < infty$ when you write $mu(Asetminus B) = mu(A)-mu(B)$. Actually, this is a stronger than what is needed: For $Bsubseteq A$, it suffices that $mu(B) < infty$. This is easy to see from the proof: we have $A = (Asetminus B) cup B$, and the two sets on the right hand side are disjoint, so we have $$mu(A) = mu(Asetminus B) + mu(B).$$
Now in order to be able to subtract $mu(B)$ from both sides of this equation, you need $mu(B) < infty$. Of course, the most natural assumption that guarantees this for arbitrary measurable sets is $mu(X) < infty$.