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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 outer measure. $E$ is a bounded set . Please could you verify the proof given here?

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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  • $\begingroup$ So, what is your question? $\endgroup$ – J. David Taylor May 7 '16 at 18:41
  • $\begingroup$ Just verify the proof. Please tell if you find a flaw. $\endgroup$ – ibnAbu May 7 '16 at 18:49
  • $\begingroup$ I don't understand what you are doing, if you consider $E_n = [-n,n]$ you won't find $i,j$ such that $\mu^*(E_i)-\mu^*(E_j)$ is as small you want $\endgroup$ – reuns May 7 '16 at 19:00
  • $\begingroup$ @user2952009. sorry I forgot to mention the set $E$ is bounded. $\endgroup$ – ibnAbu May 7 '16 at 19:11
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The result in your question is a special case for $\mathbb{R}$ of a general result.

Let $(X,\mathcal{F}, \mu)$ a measure space. Let $(E_n)$ be a sequence of subsets (measurable or not) of $X$, such that, for all $n\in\mathbb{N}$, $E_n\subseteq E_{n+1}$ and let $E=\bigcup_{n\in \mathbb{N}} E_n$. Then $$\lim_ {n\to \infty} \mu^*(E_n) = \mu^*(E) $$ where $\mu^*$ be the outer measure induced by $\mu$.

Proof: Since, for all $n\in \mathbb{N}$, $E_n \subseteq E$ and $E_n \subseteq E_{n+1}$, we have that $(\mu^*(E_n))$ is a non-decreasing sequence of non-negative real numbers bounded above by $(\mu^*(E))$. So $\lim_ {n\to \infty} \mu^*(E_n)$ exists and $$\lim_ {n\to \infty} \mu^*(E_n) \leqslant \mu^*(E) $$

Now suppose $$\lim_ {n\to \infty} \mu^*(E_n) < \mu^*(E) $$ Then, for all $n\in\mathbb{N}$, $\mu^*(E_n)<+\infty$ and so, for all $n\in\mathbb{N}$, there is $B_n$ a measurable cover of $E_n$, that is: $B_n \in \mathcal{F}$, $E_n\subseteq B_n$ and $\mu^*(E_n)=\mu(B_n)$.

Now, for all $n\in\mathbb{N}$, let $D_n = \bigcap_{i \geqslant n}B_i$. Clearly $D_n\in \mathcal{F}$. Since , for any $i \geqslant n$, $E_n\subseteq E_i\subseteq B_i$, it is easy to see that, $$ E_n\subseteq D_n = \bigcap_{i \geqslant n}B_i \subseteq B_n$$ and so $$\mu^*(E_n)=\mu(D_n)=\mu(B_n)$$ We also have that, for all $n\in\mathbb{N}$, $D_n\subseteq D_{n+1}$. Let $D=\bigcup_{n\in \mathbb{N}} D_n$. Then we have

$$\lim_ {n\to \infty} \mu^*(E_n) = \lim_ {n\to \infty} \mu(D_n)=\mu(D) < \mu^*(E) $$

On the other hand, since for all $n\in\mathbb{N}$, $ E_n\subseteq D_n $, we have $E=\bigcup_{n\in \mathbb{N}} E_n \subseteq \bigcup_{n\in \mathbb{N}} D_n=D$, so $$\mu^*(E)\leqslant \mu^*(D)=\mu(D)$$ Contardiction. So we have proved that $$\lim_ {n\to \infty} \mu^*(E_n) = \mu^*(E) $$

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  • $\begingroup$ The idea is the same : enclosing {$E_n$} with a measurable set {$B_n$ } and noticing the theorem is valid for {$B_n$}. $\endgroup$ – ibnAbu May 8 '16 at 8:29

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