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?
