Some measure Theory

**davidmccormick** · April 23rd 2010, 10:42 AM

1. Let $S = \{x \in[0,1]:$ no 4s occur in any decimal expansion of $x\}$ . Find $m^{*}(S)=0$ . I think the answer is 0, but have no idea how....

2. Given a sequence of subsets of $\mathbb{R}$ , say $\{A_i\}_{i=1}^{\infty}$ give an example to show that lim sup A_n and lim inf A_n are different.

3. I don't understand South Park's argument below. Could anyone please explain? (the <---- implication)

thanks

Originally Posted by southprkfan1

Question: $A_{\epsilon} \leq E \leq B_{\epsilon}$ makes no sense since they are sets. Do you mean they are subsets or their measures are less than or equal? It looks like you mean subsets, so I'll use that, but the proof is nearly the same either way.

-->Suppose E is Measurable

If E is measurable, then there exists a G-delta set, G and a F-sigma set, F such that $F \subseteq E \subseteq G$ and $m(G \setminus F) = 0$ . Thus, let $A_{\epsilon} = F$ and $B_{\epsilon} = G$

<--Suppose for any

there are lebesgue measurable sets

such that

and that $m(B_{\epsilon} \setminus A_{\epsilon})<\epsilon$ .

Let $A = \bigcap_{n=1}^{\infty}A_n$ and $B = \bigcap_{n=1}^{\infty}B_n$

Clearly, $A \subseteq E \subseteq B$ and A and B are both measurable.

And $m(B \setminus A) \leq 1/n$ for all n; thus, $m(B \setminus A) = 0$ .

So E differs by A (or B) by a zero set, so E is measurable.

**southprkfan1** · April 23rd 2010, 08:39 PM

Originally Posted by davidmccormick

1. Let $S = \{x \in[0,1]:$ no 4s occur in any decimal expansion of $x\}$ . Find $m^{*}(S)=0$ . I think the answer is 0, but have no idea how....

2. Given a sequence of subsets of $\mathbb{R}$ , say $\{A_i\}_{i=1}^{\infty}$ give an example to show that lim sup A_n and lim inf A_n are different.

3. I don't understand South Park's argument below. Could anyone please explain? (the <---- implication)

thanks

1. I'll try to get back.

2. Let $A_n$ = {1} if n is even and {-1} if n is odd, then limsup = 1, liminf = -1

3. What exactly are you having trouble with? I admit I skipped a few steps at the end, and the 1/n should just be an epsilon, if that helps.

**davidmccormick** · April 24th 2010, 08:30 AM

Well what are the sets, $A$ and $B$ ? Are they just the $A_{\epsilon}$ and $B_{\epsilon}$ given in the question? because then the next 2 lines are just rewriting the statement of the problem.....

**southprkfan1** · April 24th 2010, 08:39 AM

Originally Posted by davidmccormick

Well what are the sets, $A$ and $B$ ? Are they just the $A_{\epsilon}$ and $B_{\epsilon}$ given in the question? because then the next 2 lines are just rewriting the statement of the problem.....

Ok I see the problem now, it should read

$A = \bigcap_{n=1}^{\infty}A_{1/n}$ $B = \bigcap_{n=1}^{\infty}A_{1/n}$

With the rest as it was before.

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