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 correct. Who can point out where are my mistakes?
Best Answer
The above is valid because if $b_j\le b$ for $j=1,2,3,\ldots$ then $\lim_j b_j \le b$.
The above is not valid. You claim to have "taken the limit" of $m\left( \bigcup_{j=1}^N A_j \right)$. You cannot take a limit merely by putting $\infty$ wherever you see $N$. The question is: how do you know that $$ \lim_{N\to\infty} m\left( \bigcup_{j=1}^N A_j \right) = m\left( \bigcup_{j=1}^\infty A_j \right) \text{ ???} $$
First, notice that $\bigcup_{j=1}^\infty A_j$ is not defined as a limit as $N\to\infty$ of $\bigcup_{j=1}^N A_j$. Rather, it is defined by saying $x\in \bigcup_{j=1}^\infty A_j$ if and only if $\exists j\in\{1,2,3,\ldots\}\quad x\in A_j$. And if it were defined as a limit, there would still be the question of continuity of the function $m$. How would you prove that?
Here's a different way: \begin{align} m\left( \bigcup_{j=1}^\infty A_j \right) & = m \left( \bigcup_{j=1}^\infty \left( A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right)\right) & & \text{(Think about why this is true.)} \\[10pt] & = \sum_{j=1}^\infty m\left( A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right) & & \text{by countable additivity of $m$} \\[10pt] & = \lim_{N\to\infty} \sum_{j=1}^N m\left( A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right) \\[10pt] & = \lim_{N\to\infty} m\left( \bigcup_{j=1}^N A_j\setminus( A_1\cup\cdots\cup A_{j-1} ) \right) & & \text{by finite additivity of $m$} \\[10pt] & = \lim_{N\to\infty} m(A_N). \end{align}