0
$\begingroup$

Let $B$ be a subset of $[0,1]$. Suppose closure of $B = [0,1]$, prove that $m^*(B)=1$.

Proof: Clearly, I would have to show that $m^*(B)\geq 1$ and $m^*(B)\leq 1$ to imply "equal" inequality. I already showed that $m^*(B)<1$. But for $m^*(B)>1$, I only have the following so far: By countable sub-additivity, $$m^*(B\cup([0,1]\smallsetminus B) \leq m^*(B) + m^*([0,1]\smallsetminus B)$$ $$\iff$$ $$m^*([0,1]) \leq m^*(B)+m^*([0,1]\smallsetminus B)$$ $$\iff$$ $$ 1 \leq m^*(B)+m^*([0,1]\smallsetminus B).$$

Now, I just have to show that $m^*([0,1]\smallsetminus B) = 0$...but I don't know how to do that because if $B=Q$ or $B=I$, then this would have been really easy because we can use the countable property. But since we are only told that $B$ is a subset of $[0,1]$, what are the other facts or property I can use to finish the proof.

Any help would be appreciated. Thanks! :)

  • 0
    Taking B= $\mathbb{Q}$, you have that $m(B)=0$. But $B$ is dense in $[0,1]$. The statement is wrong. – 2017-01-29
  • 0
    @F.Conrad Sorry, I didn't really get what you meant. Isn't Q is dense in [0,1]? How is denseness related to outer measure in this case? – 2017-01-29
  • 0
    Ah I didnt see the tag at first. It looked like "just" a measure not an outer measure. In that case, the statement might be right – 2017-01-29
  • 0
    @learnmore Would you have some hints for me? :) – 2017-01-29
  • 0
    See the other answer. Its the same as my comment. The rationals are dense in [0,1], but their outer measure is 0. – 2017-01-29

1 Answers 1

1

For measurable set X, the outer measure of X and the measure of X must coincide. In this case, the set of rational number(measurable) in [0,1] has measure(also outer measure) 0. Thus, I think the statement is not true.

  • 0
    Did you mean the proof question was not correct, to begin with? – 2017-01-29
  • 0
    If A=Q or A=I (set of irrationals), would the statement hold? (please take a look at Question 3, https://www.math.upenn.edu/~ryrogers/HW8-solutions.pdf) – 2017-01-29
  • 0
    Your statement doesnt hold for any dense subset. That is the problem. Your question asks for ANY dense subset, but thats wrong. – 2017-01-29
  • 0
    @F.Conrad Hey Conrad, thanks for the reply. I accept this, and understand it but what I don't get was that for Question 3 with the link I posted, they proved that m*([0,1]]-Q) = m*([0,1]n(Irrationals))=1. So what you said contradicts with what they proved, assuming that the solution (UPENN) was correct.... :O – 2017-01-29
  • 0
    Everything they did was 100% correct. Its just your question here that was misleading. – 2017-01-29
  • 0
    @F.Conrad Thanks, I finally got it! I found that Irrationals are not countable so the question in the link posted was correct! – 2017-01-29
  • 0
    @F.Conrad I was thinking if the inverse would be true? "If m*(A)=1, then closure of A = [0,1]" – 2017-01-29