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Let $E_1\subset E_2\subset E_3\dots$ be subsets of $\mathbb{R}$. Prove that $m^*\big{(}\bigcup E_i\big{)}=\displaystyle{\lim_{n\to\infty}m^*(E_n)}$, where $m^*$ denotes the outer Lebesgue measure.

Okay, this is really puzzling me. I'm trying to use the regularity of the measure $m^*$ but I can't seem to find the proper way. It's obvious that $\displaystyle{\lim_{n\to\infty}m^*(E_n)}\leq m^*(\cup E_i).$ Now for the other inequality, let $\varepsilon>0$. Then for each $n\in\mathbb{N}$ there exists an open set $Α_n$ such that $E_n\subset A_n$ and $m(A_n)\leq m^*(E_n)+\varepsilon$. So we have that for each $n$, $\displaystyle{\bigcup_{i=1}^{n}E_i\subset A_n}$.

Now what though? I would appreciate any help.

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Let $\epsilon > 0$. Define the $A_n$'s such that $E_n \subset A_n$ and \begin{align*} m(A_n) \leq m^*(E_n) + \frac{\varepsilon}{n} \end{align*} Now define $G_n$ as follows. \begin{align*} G_n \doteq \cap_{k > n} A_k \end{align*} Then $\{G_n\}$ is an increasing sequence of sets, and $\cup_{n=1}^{\infty} \cap_{k > n} A_k$ is the limit set. We also have that $E_n \subset G_n \subset A_n$ since $E_n \subset A_n$ and $E_n$ are increasing.

By continuity of the premeasure, \begin{align*} m^*(\cup E_n) &\leq m^*(\cup G_n) \\ &= m(\cup G_n) \\ &= \lim m(G_n) \\ &\leq \lim m(A_n) \\ &\leq \lim m^*(E_n) + \frac{\varepsilon}{n}\\ &= \lim m^*(E_n) \end{align*}

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  • $\begingroup$ This is perfect. Thanks a lot, I find your solution very smart! $\endgroup$ – JustDroppedIn Jun 4 '17 at 22:29
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The theorem is true for bounded measurable sets.

Proof for the general case: bounded sets

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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