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If $A$ is a Lebesgue measurable subset of $[0,1]$ such that $\lambda(A)>0$. Show that there are Lebesgue measurable subsets $A_{s},0\leq s \leq \lambda (A)$, so that

(a) $A_{r}\subseteq A_{s}$ if $r\leq s$,

(b) $\lambda (A_{s})=s$ for any $0 \leq s \leq \lambda (A)$

Here is what I have at the moment...

I considered $A_{s}=A \cap [0,s]$, then part (a) is quite easy to verify.

But I only have (b) if $[0,s]\subseteq A$. My construction of the set $A_{s}$ doesn't work in general. The set $A$ can be like for example $A=[0.5,1]$, then for any $s \leq \lambda(A)=0.5$, then I have $\lambda (A_{s})=0$.

I don't quite understand the $s$ in the subscript of $A$. Have I misunderstood the problem? Or I should construct another $A_{s}$?

Thanks

Andrés E. Caicedo
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LanaDR
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  • Seeing as your $A_s$ doesn't work, I think you should construct another $A_s.$ – bof Oct 03 '17 at 01:05
  • Nothing about your construction uses the measurability of $A$. I think you should think more about what it means for $A$ to be measurable – Exit path Oct 03 '17 at 01:07
  • You're indexing the sets by real numbers, so we'll have $2^{\aleph_0}$ sets. Intuitively it seems to me that since $0\leq s/\lambda(A)\leq 1$, we're trying to chop $100s/\lambda(A)$ percent of $A$' measure, like a progress bar. No idea of how to do that so far, though. – Ivo Terek Oct 03 '17 at 01:10
  • The difficulty is that we don't know how is the measure "distributed" around $A$ (Lebesgue measurable sets can be quite crappy) – Ivo Terek Oct 03 '17 at 01:12
  • I used the fact that $A$ is measurable in my construction of $A_{s}$ because it is required to be measurable. Since $A_{s}=A \cap [0,s]$ and $A$ is measurable and $[0,s]$ is measurable and intersection of measurable sets is measurable, $A_{s}$ is measurable. – LanaDR Oct 03 '17 at 01:29

1 Answers1

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The function $f:[0,1]\to [0,\lambda(A)]$ defined by $f(x) = \lambda([0,x]\cap A)$ is continuous and nondecreasing. We have $f(0) = 0, f(1) = \lambda (A).$ By the intermediate value theorem, $f$ takes on every value in $[0,\lambda(A)].$ Thus for each $s\in [0,\lambda(a)],$ the set $f^{-1}(\{s\})$ is nonempty. And by continuity, $f^{-1}(\{s\})$ is compact. For each $s\in [0,\lambda(a)],$ we can define $x_s= \inf f^{-1}(\{s\}).$ (That way we avoid the axiom of choice.) Then $A_s= [0,x_s]\cap A$ solves your problem.

zhw.
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  • Yes. $f$ is nondecreasing implies (a) but how about (b)? How does this help for the requirement in (b)? – LanaDR Oct 03 '17 at 01:30
  • @LanaDR By surjectivity, for all $s \in [0, \lambda(A)]$ you can choose $x \in [0,1]$ so that $\lambda([0,x] \cap A)=s$. – Exit path Oct 03 '17 at 01:47
  • @leibnewtz I see. I guess I misinterpret the question. Correct me if I am wrong, so you mean like (b) is the requirement that we got to have the "onto" right? – LanaDR Oct 03 '17 at 02:08
  • @LanaDR I changed my hint to a more complete answer. – zhw. Oct 03 '17 at 02:40
  • @zhw. By the way, using the fact that the infimum of $f^{-1}(s)$ exists assumes the axiom of choice, I'm pretty sure. There's hardly ever a way to avoid it – Exit path Oct 03 '17 at 03:31
  • @leibnewtz Not that I can see. The reals with the completeness property don't need AOC for its construction. – zhw. Oct 03 '17 at 19:08