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.

  • $\begingroup$ Sorry Jack; many questions phrased similarly are homework. $\endgroup$ – Hans Parshall Oct 7 '11 at 18:52
  • $\begingroup$ I know this question is old, but it is the first result when searching for this topic. Why can we assume in this proof that $m\left(\bigcup G_n\right) = \sum m\left(G_n\right)$, just because the $G_i$ are pairwise disjoint? $\endgroup$ – mb7744 Dec 11 '14 at 15:56
  • $\begingroup$ Because in general $m\left(\bigcup G_n\right) \neq \sum m\left(G_n\right)$, in case of 2 sets it is less big expression, it is: $ m\left(G_1 \cup G_2\right)= m\left(G_1\right) + m\left(G_2\right) - m\left(G_1\cap G_2\right)$.So if $G_1 \cap G_2 = 0$ then we get what we want. $\endgroup$ – Ievgenii Nov 3 '15 at 15:55

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 $m(\bigcup_{n = 1}^\infty F_n) \leq \sum_{n = 1}^\infty m(F_n)$.

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

This link gives a proof without using monotonicity mentioned above. It is derived from the sense of "giving a room of $\epsilon$".



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