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?
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 $\alpha\in I_{n_2}$.