12
$\begingroup$

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


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 $m\left(\bigcup_{j=1}^N A_j\right)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$. Take the limit, we will have $m\left(\bigcup_{j=1}^\infty A_j\right) \leq \lim_{j\to\infty} m(A_j)$.

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

  • 0
    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...2012-04-22
  • 4
    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?2012-04-22
  • 5
    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.2012-04-22
  • 1
    Hint: Write the infinite union as a disjoint union, and use countable additivity.2012-04-22
  • 0
    @MikeB: I think the OP knows the standard proof; that seems clear from the wording in his question. :)2012-04-22
  • 0
    @cardinal: Sorry: I didn't see your comment when I was writing mine (went away half-way through, came back, finished it, and posted; then yours "appeared"). My apologies if it seemed I was stepping on your toes on purpose.2012-04-22
  • 1
    @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.2012-04-22
  • 0
    Dear @cardinal ,thanks for your help!I think you point out the key!2012-04-22
  • 0
    Dear @ArturoMagidin ,thank you too.You also point out the key!2012-04-22

2 Answers 2