# Continuity together with finite additivity implies countable additivity

I'm working through chapter 2 of Royden (4th ed.).

The problem is

Show that continuity together with finite additivity of measure implies countable additivity of measure.

My proof:

Let $\{F_k\}_{k=1}^\infty$ be a countable collection of disjoint measurable sets and define $E_n$ by $$E_1=F_1$$ $$E_k=E_{k-1}\cup F_k$$ for $k\geq 2$. So then $E_k\subset E_{k+1}$, where each $E_i$ is measurable. By construction we have, $$\bigcup_{k=1}^\infty F_k=\bigcup_{k=1}^\infty E_k.$$ Also, since the sets $F_i$ are disjoint, $$m(E_k)=m(F_1)+\dots+m(F_k).$$ Thus we have, \begin{align} m\left(\bigcup_{k=1}^\infty F_k\right)&=m\left(\bigcup_{k=1}^\infty E_k\right) \nonumber \\ &=\lim_{k\to\infty} m(E_k) \nonumber \\ &=\lim_{k\to\infty} m(F_1)+\cdots+m(F_k)\nonumber\\ &=\sum\limits_{k=1}^\infty m(F_k) \nonumber \end{align}

I've been working through this book and noticed that construction is used quite a bit and I'm wondering if anyone could give me any insight on determining how to construct the right set. I figured out the one above by trial and error with some reasoning, but I don't really have a clear approach.

• Your proof looks correct to me. It is indeed the case that one often passes from disjoint sets to their successive unions and also from sets included in each other to their successive differences (which are then disjoint). Disjoint sets are useful because of additivity while the inclusions (both upward and downward) because of continuity. Which form is better depends on the problem. I don't think there's much more to it than that. – Marek Jul 21 '13 at 7:30
• I see. I guess there isn't a clean cut answer afer all, just a matter of context. Thank you for the response. – Shant Danielian Jul 21 '13 at 7:36

## 1 Answer

It seems that proof going in the aforementioned direction required converting disjoint sets to increasing sets.

Proving continuity given countable additivity requires converting increasing sets to disjoint sets.

I have a feeling that proofs such as these involve converting disjoint to increasing, increasing to decreasing, decreasing to disjoint and vice-versa.

Given continuity of measure, prove countable additivity to prove measure

Edit: Oh it looks like @Marek Ishtar gave the answer already "Disjoint sets are useful because of additivity while the inclusions (both upward and downward) because of continuity. Which form is better depends on the problem. I don't think there's much more to it than that."