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.
Best Answer
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.