Consider the Cantor set $C$. Define a new "Cantor" set which we will call $C_\alpha$ for $\alpha \in (0,1)$. We construct $C_\alpha$ by letting $C_0 = [0,1]$, $C_1 = C_0$ but with an open interval of length $\alpha$ taken out of the center. Define $C_2$ by taking $C_1$ and taking out of each remaining half an interval of length $b \cdot \alpha$ where $b$ is the length of one of the remaining halves in $C_2$. Continue in this manner for all sets $C_k$ with $k \in \mathbb{N}$. Then we create our new "Cantor" set $C_\alpha$ by taking $C_\alpha = \displaystyle\bigcap_{i=0}^\infty C_i$.
I seek to show that the complement of $C_\alpha$ in the unit interval $[0,1]$ has measure $1$ (Notation: when I write $C_i^c$, I mean with respect to $[0,1]$, not all of $\mathbb{R}$). First, note that $|C_0^c| = 0, |C_1^c| = \alpha, |C_2^c| = 2\alpha^2$. In general, we will see that $|C_k^c| = 2^{k-1}\alpha^k$. Then I obtain
$$\displaystyle\sum_{i=0}^\infty |C_i^c| = \displaystyle\sum_{i=0}^\infty 2^{k-1}\alpha^k = \frac{1}{2} \displaystyle\sum_{i=0}^\infty (2\alpha)^k = \dots ?$$
At this point, I get stuck because I want to eventually use $\displaystyle\sum_{k=0}^\infty x^k = \frac{1}{1-x}$ for $|x| < 1$ since I suspect that is the way to do this, but I cannot be sure that $|2\alpha| < 1$. In fact, this won't be true for all $\alpha \in (0,1)$ whenever $\alpha \geq \frac{1}{2}$. Any suggestions on where I should go? I'm trying to show that $\displaystyle\sum_{i=0}^\infty |C_i^c| =1$.
First of all, we note that $C_1\supset C_2\supset\cdots$, hence $C^c=\bigcup_{k=0}^\infty C_k^c$ is the union of a sequence of increasing sets. So $$m(C^c)=\lim_{k\to\infty} m(C_k^c) \qquad \text{(rather than the summation).}$$
Secondly, your computation of $m(C_k^c)$ is incorrect. After $k-1$ step, the total length of the remaining part is $m(C_{k-1})$, so in the $k$ step, we drop several intervals of total length $\alpha\cdot m(C_{k-1})$ out of the remaining part after $k-1$ step. Hence $$m(C_k^c)=m(C_{k-1}^c)+\alpha\cdot m(C_{k-1})=\alpha+(1-\alpha)m(C_{k-1}^c). \tag{1}$$ Let me use more words to describe the procedure: Assume that after $k-1$ step, the remaining part is $C_{k-1}=I_1\cup I_2\cup \cdots \cup I_p$ (disjoint union); then in the $k$ step, we drop an interval of length $\alpha\cdot |I_i|$ from the interval $I_i$, hence $$m(C_k)=m(C_{k-1})-\sum_{i=1}^p\alpha|I_i|=m(C_{k-1})-\alpha\cdot m(C_{k-1})\tag{2}.$$ Hence $(1)$ follows. Combining $(1)$ with $m(C_1^c)=\alpha$, or combining $(2)$ with $m(C_1)=1-\alpha$, we obtain $$m(C_k^c)=1-(1-\alpha)^k, \qquad k\geq1.$$
Therefore $$m(C^c)=\lim_{k\to\infty} m(C_k^c)=1.$$