I’m going to assume that Cantor set here refers to the standard middle-thirds Cantor set $C$ described here. It can be described as the set of real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$, i.e., real numbers of the form $$\sum_{n=1}^\infty \frac{a_n}{3^n},$$ where each $a_n$ is either $0$ or $2$.
For each positive integer $n$ let $D_n = \{0,1\}$ with the discrete topology, and let $$X = \prod_{n=1}^\infty D_n$$ with the product topology. Elements of $X$ are infinite sequences of $0$’s and $1$’s, so $(0,0,0,1)$ and $(0,1,1,1,1,1,1)$ are not elements of $X$; if you pad these with an infinite string of $0$’s to get $(0,0,0,1,0,0,0,0,\dots)$ and $(0,1,1,1,1,1,1,0,0,0,0,\dots)$, however, you do get points of $X$. A more interesting point of $X$ is the sequence $(p_n)_n$, where $p_n = 1$ if $n$ is prime, and $p_n = 0$ if $n$ is not prime.
Your problem is to show that $C$, with the topology that it inherits from $\mathbb{R}$, is homeomorphic to $X$. To do that, you must find a bijection $h:C\to D$ such that both $h$ and $h^{-1}$ are continuous. The suggestion that you found is to let $$h\left(\sum_{n=1}^\infty\frac{a_n}{3^n}\right) = \left(\frac{a_1}2,\frac{a_2}2,\frac{a_3}2,\dots\right).$$ Note that $$\frac{a_n}2 = \begin{cases}0,&\text{if }a_n=0\\1,&\text{if }a_n=2,\end{cases}$$ so this really does define a point in $X$. This really is a bijection: if $b = (b_n)_n \in X$, $$h^{-1}(b) = \sum_{n=1}^\infty\frac{2b_n}{3^n}.$$
Yes, it is.
Note that the Cantor set is in fact a compact group, and the product measure $\mu$ is the Haar measure, that is, the invariant finite, regular Borel measure, which is by Haar's theorem unique up to a scaling factor.
That Lebesgue measure restricted to the fat Cantor set is invariant is an easy exercise (it follows easily from the fact that the Lebesgue measure is translation invariant).
It is also regular, because it is a restriction of a regular measure, so by Haar's theorem it is a scalar multiple of the product measure.
Best Answer
For any $x$ in the Cantor set $C$ there is a unique $f(x)= (x_n)_{n\in N}\in \{0,1\}^N$ such that $\sum_{n=1}^{\infty} 2 x_n 3^{-n}=x.$ For $j\in N$ let $H(j)$ be the set of functions from $\{1,...,j\}$ to $\{0,1\}$. For $a\in \{0,1\}$ and $j\in N$ let $H(j,a)=\{h\in H(j):h(j)=a\}$. We have $$f^{-1}U(j,a)=\{x\in C: x_j=a\}=\cup_{h\in H(j,a)}\{x\in C: \forall i\leq j (x_i=h(i))\}.$$ Observe that for $a=0$ and $j\in N$ and $h\in H(j,0)$ we have $$\{x\in C:\forall i\leq j\, (\,x_i=h(i)\,)\}= C\cap [h^*,h^*+2.3^{-j})$$ $$\text {where } h^*=\sum_{i=1}^{i=j} 2 h(i) 3^{-i}.$$ Now either $h^*=0$ or $h^*$ is the upper endpoint of the open interval $(h^*-3^{-(j-1)},h^*)$ that was removed in the construction of $C$. Therefore, when $a=0$, $$\{x\in C:\forall i\leq j (x=h(i)\}=C\cap (h^*-3^{j-1},h^*+2.3^{-j})$$ which is open in $C$. The case $a=1$ is handled similarly.