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Let $\bigcup_{n=1}^\infty E_n=E$ and $ E_{n} \subseteq E_{n+1} $ then $\lim\limits_{n\mapsto \infty} \mu^*(E_n) = \mu^*(E) $ even if each $E_n$ is a non-measurable set, where $\mu^*$ is Lebesgue outer measure. $E$ is a bounded set.Proof sketch please?

This theorem allows the short proofs of Dominated convergence theorem, Vitali Convergence Theorem, Monotone Convergence Theorem , Egorov's theorem and Luzin's theorem without dwelling much on the machinery of measure theory.

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  • 1
    $\begingroup$ The proof is straight forward for measurable sets but not so for arbitrary sets. Actually it is true and it was given as an exercise in 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 )' by Russell A. Gordon $\endgroup$ – ibnAbu May 5 '16 at 0:51
  • $\begingroup$ It is theorem 1.15 in the book mentioned above. $\endgroup$ – ibnAbu May 5 '16 at 11:16
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The theorem is true for measurable sets.

Proof for the general case:

There is a subsequence { $ E_k $} such that $\mu^* (E_{k+1} ) - \mu^* (E_k ) \le \frac {\epsilon}{2^{k+1}} $

Lets first construct such subsequence :

$ \lim\limits_{n\mapsto \infty}\mu^*(E_n) \ge \mu^*(E_{n+1}) \ge \mu^*(E_n)$

Choose $E_1$ such that $ \lim\limits_{n\mapsto \infty}\mu^*(E_n) -\mu^*(E_1) \le \frac {\epsilon}{2} $

Choose $E_2$ such that $E_1 \subseteq E_2$ and $ \lim\limits_{n\mapsto \infty}\mu^*(E_n) -\mu^*(E_2) \le \frac {\epsilon}{2^3} $

Choose $E_3$ such that $E_2 \subseteq E_3$ and $ \lim\limits_{n\mapsto \infty}\mu^*(E_n) -\mu^*(E_3) \le \frac {\epsilon}{2^4} $ Then use induction

Step 1: cover $E_k$ with union of open intervals $\bigcup_{i=1}^\infty I_i = L_k $ Such that $ \mu^* (L_k) \le \mu^* (E_k) + \frac {\epsilon}{2^k} $

By Caratheodory condition $ \mu^* (E_{k+1} \bigcap L_k^c ) = \mu^* (E_{k+1} ) - \mu^* (E_{k+1} \bigcap L_k ) $ $\mu^*(E_k) \le \mu^*(E_{k+1} \bigcap L_k ) \le \mu^*(L_k) $

Therefore $ \mu^*(E_{k+1} \bigcap L_k^c ) \le \mu^*(E_{k+1}) - \mu^* (E_k) \le \frac {\epsilon}{2^{k+1}}$

Step 2 : Let $ G_{k+1} = E_{k+1} \bigcap L_k^c $

$ L_k \bigcup G_{k+1}$ contains $E_{k+1}$

Now cover $ L_k \bigcup G_{k+1}$ with union of intervals $\bigcup_{i=1}^\infty I_i = H_{k+1} $ Such that $\mu^* (H_{k+1}) \le \mu^*( L_k \bigcup G_{k+1}) + \frac {\epsilon}{2^{k+1}} $

$\mu^* (H_{k+1}) \le \mu^*( L_k) + \mu^*( G_{k+1}) \le \mu^*(E_k) + \frac { \epsilon}{2^k} +\frac { 2\epsilon}{2^{k+1} }$

Now using $H_{k+1} $ as the cover for $ E_{k+1} $

As seen $\mu^*(E_{k+1}) \le \mu^*(H_{k+1})$ $\le \mu^*(E_{k+1})$ $+ \frac {\epsilon}{2^k}$ $+\frac {2\epsilon}{2^{k+1}} $

Now apply step 1 and 2 to $ E_{k+1}$ and $ E_{k+2}$ and get :

$ \mu^*(E_{k+2}) \le \mu^* (H_{k+2}) \le \mu^*(E_{k+2}) + \frac { \epsilon}{2^k} +\frac { 2\epsilon}{2^{k+1} } +\frac { 2\epsilon}{2^{k+2} }$

It is obvious $ E \subseteq \bigcup_{i=1}^\infty H_k $

$H_k \subseteq H_{k+1}$

$ \mu^*(E_{k}) \le \mu^* (H_{k}) \le \mu^*(E_{k}) + 4 \epsilon $

Notice that the theorem is valid for $H_k$ as it is a measurable set (union of intervals) So $\lim\limits_{k\mapsto \infty} \mu^*(H_k) \ge \mu^*(E) $

$\lim\limits_{k\mapsto \infty}\mu^*(E_{k}) \le \lim\limits_{k\mapsto \infty} \mu^*(H_k) \le \lim\limits_{k\mapsto \infty}\mu^*(E_{k}) + 4\epsilon $

$\lim\limits_{k\mapsto \infty}\mu^*(E_{k}) \le \mu^*(E) \le \lim\limits_{k\mapsto \infty}\mu^*(E_{k}) + 4\epsilon $

Because $\epsilon $ is arbitrary the proof is complete.

Remark: The proof is straight forward for measurable sets but not so for arbitrary sets. Actually it is true and it was given as an exercise in 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 )' by Russell A. Gordon . It is theorem 1.15 in the book.

This theorem allows the short proofs of Dominated convergence theorem, Vitali Convergence Theorem, Monotone Convergence Theorem , Egorov's theorem and Luzin's theorem without dwelling much on the machinery of measure theory.

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  • 3
    $\begingroup$ It is also exercise 4 of section 12 in Halmos' Measure theory. It is not true for all outer measures, but it is for Lebesgue measure because it is regular. You should precise that $\mu$ is Lebesgue measure in your question. $\endgroup$ – Sergio May 13 '16 at 20:35

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