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I am going to try and prove that the outer measure of $[0, 1] $ is $0$. I would be grateful if someone could point out the mistake.

The outer measure of an interval is defined as $\inf \Sigma {l (I_n)} $ for any open covering of the interval $\{ I_n\} $. Procedure: Form a injection between $Q\cap [0, 1] $ and $N $. Around each rational number in $Q\cap [0, 1] $, create a neighbourhood of radius $\frac {1}{2^n} $ where $n $ is the natural number to which this rational nunber is mapped. This is a covering of $[0, 1] $, and the sum of the lengths of the open sets, which is 1, is greater than the outer measure.

Next around the same rational points, create intervals of length $\frac {1}{2^{n+1}} $. Now the sum of lengths is $1/2$, which is still greater than the outer measure by definition.

Going on like this, we can show that the outer measure is equal to 0. Where am I going wrong?

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    Wny do you think that sequence $I_n$ covers all irrational elements of $[0,1]$? It just shows that, for each $k$, here are some real numbers $\alpha$ such that $|\alpha-r_n|>\frac{1}{2^{n+k}}$, where $r_n$ is the $n$th enumerated rational in $[0,1]$. – Thomas Andrews Aug 29 '15 at 20:12
  • @ThomasAndrews-I'm creating neighbourhoods around all rational points. Shouldn't such neighbourhoods contain all irrational numbers also? –  Aug 29 '15 at 20:13
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    No. You'd have to prove that. And it is not true. It's easy enough to construct such an $\alpha$ not in $\bigcup I_n$ - just use the same argument as Cantor, very roughly. – Thomas Andrews Aug 29 '15 at 20:14
  • $[0,1]\setminus{\frac1\pi}$ contains a neighborhood around all rational points, and isn't equal to $[0,1]$. (Your set is a much more serious offender.) – Akiva Weinberger Aug 30 '15 at 05:50

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It is counter-intuitive, but your $I_k$ do not cover $[0,1]$. It was a thing that bugged me for a while, too.

Let $\alpha_n$ be a real number not in $\bigcup_{k=1}^{n} I_k$. Then some sub-sequence of $\{\alpha_n\}$ must converge in $[0,1]$. Let $\alpha$ be that limit. Then show that $\alpha$ is not in $I_n$ for any $n$.

You can see this as a result of the fact that $[0,1]$ is compact - if $\bigcup I_n=[0,1]$ then some finite subset of the $I_n$ cover $[0,1]$.

What this means is that, while there are always rationals arbitrarily close to any real $\alpha$, the enumerated rationals don't generally get to $\alpha$ "fast enough." For any $M>1$, there are a huge number of irrationals $\alpha$ such that that $\left|\alpha-r_n\right|>\frac{1}{M2^n}$ for all $n$.

For any $n_1$, we know there is an $n_2$ such that $$\left|\alpha-r_{n_2}\right|<\frac{1}{M2^{n_1}}$$ but nothing says that $n_2\leq n_1$ - that is we cannot conclude that:

$$\left|\alpha-r_{n_2}\right|<\frac{1}{M2^{n_2}}$$

which would be what you'd need to show that $\alpha\in I_{n_2}$.

Thomas Andrews
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  • How can you prove that there will always exist such an $\alpha_n $? Maybe after a particular $n $, we have a covering of the interval! –  Aug 29 '15 at 21:11
  • One may say that the sum of the measures of the sets of a covering should exceed the measure of the interval. However, nowhere in the definition of outer measure is this mentioned –  Aug 29 '15 at 21:12
  • It is true that outer measure is not countably additive, but it is sub-additive. If $\phi$ is the outer measure, then $$\phi\left(\bigcup_{i=1}^\infty A_i\right)\leq \sum_{i=1}^\infty\phi(A_i)$$ – Thomas Andrews Aug 29 '15 at 21:20
  • It's a painful exercise, but you can directly prove that the union of finitely many intervals that have lengths that do not add up to more than $1$ cannot cover the unit interval. This is basically an induction proof - the complement of a finite union of open intervals inside $[0,1]$ is a union of finitely many disjoint closed intervals. It's a tedious theorem, but not hard to prove. – Thomas Andrews Aug 29 '15 at 21:24