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, $$m\left( \bigcup_{x\in C} I_x\right) = \frac{11}{10} - \bigg( \frac{1}{3} - \frac{1}{10} \bigg) - 2\times \bigg( \frac{1}{9} - \frac{1}{10} \bigg) = \frac{76}{90}.$$
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.