# 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

Of course, as it is mentioned in the comments you were wrong in assuming $$\lim_{N\to\infty}m\left(\bigcup_{j=1}^NA_j\right)=m\left(\bigcup_{j=1}^{\infty}A_j\right)$$ However the proof can be done without using the countably additivity property. Suppose $m$ is a measure defined on $\mathcal{F}$, the field generating the $\sigma$-field $\mathcal{A}$, then for a general $A\in\mathcal{A}$, the outer measure is defined as $$m^{\ast}(A)=\inf\{m(B):B\in\mathcal{F}_{\sigma},B\supset A\}$$ We can show that $m^{\ast}$ is monotone and strongly subadditive. Now let $A_n\uparrow A$.
Then as $A_n\subset A$, so $m^{\ast}(A_n)\leq m^{\ast}(A)$. So $$\lim_{n\to\infty}m^{\ast}(A_n)\leq m^{\ast}(A)$$. Now obviously we can find $B_n\in\mathcal{F}_{\sigma}$ such that $$m(B_n)\leq m^{\ast}(A_n)+\frac{\epsilon}{2^n}$$ by the definition of $m^{\ast}$. Let $$C_n=\bigcup_{k=1}^nB_k$$. Then $A_n\subset B_n\subset C_n$ and hence $\lim A_n\subset\lim C_n$ or $m^{\ast}(A)\leq m^{\ast}(\lim C_n)$. Now $\lim C_n\in \mathcal{F}_{\sigma}$ so $m^{\ast}(\lim C_n)=m(\lim C_n)$ and as $m$ is continuous from below, so $m(\lim C_n)=\lim m(C_n)$. Now $m(C_{n+1})+m(C_n\cap B_{n+1})=m(C_n)+m(B_{n+1})$. Now note that $A_n\subset C_n$ and $A_n\subset A_{n+1}\subset B_{n+1}$. So $A_n\subset C_n\cap B_{n+1}$, so $m^{\ast}(A_n)\leq m^{\ast}(C_n\cap B_{n+1})=m(C_n\cap B_{n+1})$. Thus our claim is proved. So we have $m^{\ast}(A)\leq \lim m(C_n)\leq \lim m^{\ast}(A_n)+\epsilon$. So $$m^{\ast}(A_n)\uparrow m^{\ast}(A).$$