# Proving the measure of an increasing sequence of measurable sets is the limit of the measures

Show that if $A_1\subseteq A_2\subseteq A_3\cdots$ is an increasing sequence of measurable sets(so $A_j\subseteq A_{j+1}$ for every positive integer $j$),then we have $$m(\bigcup_{j=1}^{\infty}A_j)=\lim_{j\to\infty}m(A_j)$$

Here is my proof:

According to the $\sigma-$algebra property,$\bigcup_{j=1}^{\infty}A_j$ is a measurable set,so it makes sense to talk about $m(\bigcup_{j=1}^{\infty}A_j)$.

Firstly I prove that $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$.This is because for any given positive integer $N$,$A_N\subseteq \bigcup_{j=1}^{\infty}A_i$,according to monotonicity,we have $m(A_N)\leq m(\bigcup_{j=1}^{\infty}A_i)$.Take the limit,we will have $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$.

Secondly I prove that $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$.For any given positive integer $N$,$\bigcup_{j=1}^{N}A_j= A_N$.According to monotonicity,we have $m(\bigcup_{j=1}^{N}A_j)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$.Take the limit,we will have $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$.

Combine the above two arguments ,we will see that $$m(\bigcup_{j=1}^{\infty}A_j)=\lim_{j\to\infty}m(A_j)$$$\Box$

The above is my proof,unlike many books,my proof does not use the property of countable additivity.So I doubt my proof is false.Who can point out where are my mistakes?

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At least somewhere you should remark that $m(A_j)$ is an increasing sequence, and so the limit exists and is less than or equal to any common bound to all terms... – Arturo Magidin Apr 22 '12 at 1:55
First of all, you have $\lim_{N \to \infty} m(A_j)$ on the right-hand side in part of the second step. Should your $N$ be a $j$? Second of all, how do you get from $\lim_{N\to\infty} m(\cup_{j=1}^N A_j)$ to $m(\cup_{j=1}^\infty A_j)$ when taking the limit at the very end? – cardinal Apr 22 '12 at 2:00
In the second half, you are trying to argue that $$\lim_{N\to\infty}m(\cup_{j=1}^NA_j) = m(\lim_{N\to\infty}\cup_{j=1}^NA_j) = m(\cup_{j=1}^{\infty}A_j)$$(in order to "take the limit"). But this is precisely what you are trying to prove. – Arturo Magidin Apr 22 '12 at 2:04
Hint: Write the infinite union as a disjoint union, and use countable additivity. – Mike B Apr 22 '12 at 2:35
@ArturoMagidin: No, I knew you weren't. :) I just was wondering if my statement was a little too obtuse/Socratic. I've noticed the synchronization of how comments get posted can be strange at times. – cardinal Apr 22 '12 at 2:49