Theorem (Monotone convergence theorem for measurable sets): Let $E_1 \subset E_2 \subset \ldots \subset \mathbb{R}^d$ be a countable non-decreasing sequence of Lebesgue measurable sets. Then $$m\bigg(\bigcup_{n=1}^{\infty}E_n\bigg)=\lim_{n \to \infty} m(E_n)$$ Again let $\mathbb{R}^d \supset E_1 \supset E_2 \supset \ldots $ be a countable non-increasing sequence of Lebesgue measurable sets. If at least one of the $m(E_n)$ is finite, then $$m\bigg(\bigcap_{n=1}^{\infty}E_n\bigg)=\lim_{n \to \infty} m(E_n)$$
Using this theorem, I have to prove the following theorem:
Theorem (Dominated convergence theorem for measurable sets): Let $E_1, E_2, \ldots $ are Lebesgue measurable sets in $\mathbb{R}^d$ such that $(i)$ $E_n$ are all contained in another Lebesgue measurable set $F$ of finite measure and $(ii)$ the sequence of sets $\{E_n\}_{n=1}^{\infty}$ converges pointwise to the set $E$. Then $$\lim_{n \to \infty}m(E_n)=m(E)$$
Note that I cannot use MCT or DCT for measurable functions, or any other advanced machinery in order to avoid circularity. The book strictly says to use MCT for measurable sets only!
Note that $$\liminf E_n = \bigcup_n \bigcap_{k\ge n} E_k $$ Hence, by the MCT $$ m(\liminf E_n) = \lim_n m\left(\bigcap_{k\ge n} E_k\right) \le \liminf m(E_n) $$ On the other hand $$ \limsup E_n = \bigcap_n \bigcup_{k\ge n}E_k $$ gives by the MCT $$ m(\limsup E_n) = \lim_n m\left(\bigcup_{k\ge n} E_k\right) \ge \limsup m(E_n) $$ Hence, as $\liminf E_n = \limsup E_n = \lim E_n$ $$ m(\lim E_n) = m(\liminf E_n) \le \liminf m(E_n) \le \limsup m(E_n) \le m(\limsup E_n) = m(\lim E_n) $$ This implies $$ \liminf m(E_n) = \limsup m(E_n) = m(\lim E_n) $$ therefore $m(E_n)$ converges to $m(E)$.