Proof that the Fat Cantor set is not a null set To be specific, the Fat Cantor set (denoted $F$) in question here is the one obtained when removing an interval of length $\frac{1}{4}$ from $[0,1]$, an interval of length $\frac{1}{16}$ from the two intervals obtained from step 1, etc. Also, a set $A \subset \mathbb{R}$ is denoted a null set if for each $\epsilon > 0$ there is a countable covering of $A$ by open intervals $(a_i, b_i)$ such that $$\sum_{i=1}^{\infty} b_i - a_i \leq \epsilon$$
My proof is as follows:
The length of the $F^c$ (i.e. the intervals removed during the construction) is given by $$\sum_{n=0}^{\infty} 2^n (\frac{1}{4})^{n+1} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16}...$$ which converges to $\frac{1}{2}$. Hence, the length of $F \rightarrow \frac{1}{2}$ as $n\rightarrow \infty$, implying that $F$ has nonzero outer measure, and so is not a null set.
Did I miss or gloss over anything? To be honest this explanation seems a little too straightforward, so any comments/hole-poking would be much appreciated.
 A: You are right. Fat Cantor set is a nowhere dense set with positive Lebesgue measure. In measure theory, set with Lebesgue measure zero is called null set. 
A: There's a point that was glossed over, since you ask:
If I'm interpreting the "etc" correctly, then $K=\bigcap K_n$, where $K_n$ is the union of $2^n$ intervals of length $\delta_n$, where $\delta_0=1$ and $$\delta_{n+1}=\frac{\delta_n-4^{-(n+1)}}{2}.$$
What's missing is  a proof that $\delta_n>4^{-(n+1)}$; you need this because if not then in going from $K_n$ to $K_{n+1}$ you replace each interval by itself minus a longer interval.
It's clear that at each stage the "remaining intervals" are shorter than the remaining intervals in the construction of the middle-thirds set; indeed you can easily show $\delta_n>3^{-n}$ by induction.
Or if that's not enough fun you can show that in fact $$\delta_n=2^{-n-1}+2^{-2n-1}.$$ (Of course it's trivial to verify that; I hope someone does so just to check my algebra. If you're curious how one might find that formula: Write the recurrence as $2\delta_{n+1}-\delta_n=-4^{-(n+1)}$; now by analogy with the standard technique for first-order linear DEs write $\delta_n=2^{-n}\alpha_n$ and you get $\alpha_{n+1}-\alpha_n=$ something explicitly summable...)
