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

Solution 1:

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

The above is valid because if $b_j\le b$ for $j=1,2,3,\ldots$ then $\lim_j b_j \le b$.

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

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}

Solution 2:

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 $$m^{\ast}(A_n)\uparrow m^{\ast}(A).$$