Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\lbrace F_n \rbrace$ be a sequence of sets in a $\sigma$-algebra $\mathcal{A}$. I want to show that $$m\left(\bigcup F_n\right)\leq \sum m\left(F_n\right)$$ where $m$ is a countable additive measure defined for all sets in a $\sigma$ algebra $\mathcal{A}$.

I think I have to use the monotonicity property somewhere in the proof, but I don't how to start it. I'd appreciate a little help.

Added: From Hans' answer I make the following additions. From the construction given in Hans' answer, it is clear the $\bigcup F_n = \bigcup G_n$ and $G_n \cap G_m = \emptyset$ for all $m\neq n$. So $$m\left(\bigcup F_n\right)=m\left(\bigcup G_n\right) = \sum m\left(G_n\right).$$ Also from the construction, we have $G_n \subset F_n$ for all $n$ and so by monotonicity, we have $m\left(G_n\right) \leq m\left(F_n\right)$. Finally we would have $$\sum m(G_n) \leq \sum m(F_n).$$ and the result follows.

share|improve this question
If this is homework please tag it as such. –  Hans Parshall Oct 7 '11 at 18:47
This is not homework... –  Jack Oct 7 '11 at 18:51
Sorry Jack; many questions phrased similarly are homework. –  Hans Parshall Oct 7 '11 at 18:52
add comment

1 Answer

up vote 3 down vote accepted

Given a union of sets $\bigcup_{n = 1}^\infty F_n$, you can create a disjoint union of sets as follows.

Set $G_1 = F_1$, $G_2 = F_2 \setminus F_1$, $G_3 = F_3 \setminus (F_1 \cup F_2)$, and so on. Can you see what $G_n$ needs to be?

Using $m(\bigcup_{n = 1}^\infty G_n)$ and monotonicity, you can prove $m(\bigcup_{n = 1}^\infty F_n) \leq \sum_{n = 1}^\infty m(F_n)$.

share|improve this answer
I can instead see that $G_{n+1}=F_{n+1}$ \ $\bigcup F_k$. I'm not sure about $G_n$. –  Jack Oct 7 '11 at 19:19
What does $k$ range over? If you know a general form for $G_{n + 1}$, what is different about $G_n$? –  Hans Parshall Oct 7 '11 at 19:20
$k$ goes from $1$ to $n$. Then I suppose $G_n = F_n$\ $\bigcup F_k$ where $k$ goes from $1$ to $n-1$. right? –  Jack Oct 7 '11 at 19:24
Sure. Do you see how this helps with your problem? –  Hans Parshall Oct 7 '11 at 19:29
Yes! I do. I have made additions. would you please see if its alright? Thanks. –  Jack Oct 7 '11 at 19:38
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.