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!

  • $\begingroup$ I would assume that the definition of "converge pointwise" for sets would exactly be the condition $E=\cap E_n.$ $\endgroup$ – Leon Sot Dec 15 '16 at 10:33
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    $\begingroup$ I'm very interested to know what book you are reading. $\endgroup$ – Hua Dec 15 '16 at 16:19
  • $\begingroup$ @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). $\endgroup$ – 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 $m(E)$.

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  • $\begingroup$ Great! Thanks a lot! $\endgroup$ – user398842 Dec 15 '16 at 11:26

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