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I have a three part question i'm trying to work through.

First part is in title,

(a) Suppose that there exists a measurable set E s.t. $m(E)=1$, and from this show then there then must be a measurable $F \subset E$ s.t. $m(F)=1/2$. [Hint: consider the function $f(x) = m(E \cap (-\infty,x])$

For this part, I want to invoke the intermediate value theorem to show that there must be an x for which this function has measure $1/2$, although i'm not sure if I'm allowed to do that.

(b) There is a closed set $F$, consisting entirely of irrationals, such that $F \subset E$ and $m(F)=1/2$

For this part, my first instinct would be to use the same $F$ that I found in part (a) and then remove all the rational numbers from it, this set will still have a measure of one half because I only removed a countable subset from the original set. As far as gaurenteeing that this set is closed, I haven't the foggiest idea.

(c) There is a compact set $F$ with empty interior such that $F \subset E$ and $m(F)=1/2$

I think this part will become more clear once I have a better understanding of part (b), although gaurenteeing that the compact set $F$ that i'm looking for has an empty interior is a bit intimidating. Definitely going to need you're guys's help on this one!!

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For a) you have to observe that $|f(x)-f(y)| \leq |x-y|$ so $f$ is continuous and IVP can be used. For b) consider $[0,1]\setminus \cup I_n$ where $I_n=(r_n-\frac 1 {2^{n+1}},r_n+\frac 1 {2^{n+1}})$, $\{r_n\}$ being the set of all rationals in $[0,1]$. This set has measure $>1-\sum \frac 1 {2^{n+1}}$ so it is a set of irrationals with measure at least 1/2. Now increase the length of the first interval to get measure exactly 1/2. For c) you should look at "Cantor like sets". For every $a \in (0,1)$ there exists a Cantor like set of measure $a$.

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We can do (b),(c) together. Let $I$ be the set of irrationals. Then $m(E)= m(E\cap I) = 1.$ By the inner regularity of Lebesgue measure, there is a compact $K\subset E\cap I$ such that $m(K) > 1/2.$ The function $g(x)= m[(-\infty,x]\cap K]$ is continuous on $\mathbb R,$ with $\lim_{x\to -\infty} g(x) = 0,$ $ \lim_{x\to \infty} g(x) = m(K).$ By the IVP, $g(x_0) = 1/2$ for some $x_0.$ The set $(-\infty,x_0]\cap K$ then satisfies the requirements of both (b),(c).

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  • $\begingroup$ I have not learned about this inner regularity property yet... does it relate to the topological separation property of regularity? $\endgroup$ – Math is hard Jun 19 '18 at 21:16

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