# Dominated convergence theorem of Lebesgue measurable sets using monotone convergence theorem

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!

• I would assume that the definition of "converge pointwise" for sets would exactly be the condition $E=\cap E_n.$ – Leon Sot Dec 15 '16 at 10:33
• I'm very interested to know what book you are reading. – Hua Dec 15 '16 at 16:19
• @Hua I think it's Introduction to Measure Theory by Terry Tao. I've come across the same question (and ran into the same difficulty). – AspiringMathematician Apr 11 '17 at 15:29

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 $m(E)$.