<s>[INST] Evaluate $\displaystyle m\left( \bigcup_{x \in C} \left( x - \frac{1}{20}, x+\frac{1}{20} \right) \right)$ where $C$ is Cantor set. 
Question: Let $C$ be the Cantor set, obtained from the unit interval $[0,1].$
      For each $x \in C$, let $I_x$ be the interval $\displaystyle \left(x-\frac{1}{20}, x + \frac{1}{20}\right).$
      Note that $\displaystyle \bigcup_{x \in C}I_x$ is open. 
      Evaluate $\displaystyle m\left( \bigcup_{x \in C}I_x \right).$ 

My attempt: 
It seems to me that $\bigcup_{x\in C}I_x$ is an open cover for $C.$
Since $0\in C$ and $1\in C,$ therefore 
$$m\left( \bigcup_{x\in C}I_x \right) = 1+\frac{1}{20} + \frac{1}{20} = 1.1$$
But I am not sure whether this makes sense. 
EDITED (based on hint provided by Ross Millikan): Observe that intervals missing from $\bigcup_{x\in C} I_x$ are 
$$\bigg[\frac{1}{3} + \frac{1}{20} , \frac{2}{3} - \frac{1}{20}\bigg], \bigg[ \frac{2}{9} + \frac{1}{20}, \frac{3}{9} - \frac{1}{20} \bigg] \text{ and } \bigg[ \frac{7}{9} + \frac{1}{20} ,\frac{8}{9} - \frac{1}{20} \bigg] .$$
Therefore, 
$$m\left( \bigcup_{x\in C} I_x\right) = \frac{11}{10} - \bigg( \frac{1}{3} - \frac{1}{10} \bigg) - 2\times \bigg( \frac{1}{9} - \frac{1}{10} \bigg) = \frac{76}{90}.$$
 [/INST] Since the whole interval $(\frac 13,\frac 23)$ is excluded from the Cantor set the interval $(\frac 13+\frac 1{20},\frac 23-\frac 1{20})$ is excluded from your union.  The measure is then less than $1.1$.  You need to find all the intervals that are missing from your set, add them up, and subtract from $1.1$.  There aren't very many of them.
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