$c$ is the Cantor function, $C$ the Cantor set, and $\rho$ the Lebesgue measure
We consider $\mu$ as $\mu(A) = \rho(c(A \cap C))$ for each each element of the tribute
Is $\mu$ currently defining probability measure on the Cantor set (means the support of $\mu$ is C) ?
I have issue showing the sigma-additivity...
I assume you want to define $\mu$ on the Borel (or Lebesgue) $\sigma$-algebra in $\Bbb R$.
Let $c$ be the Cantor function, $C$ the Cantor set, and $\rho$ the Lebesgue measure. Let $\mathcal{B}$ be the Borel $\sigma$-algebra in $\Bbb R$.
Let $\mu(A) = \rho(c(A \cap C))$ for each each element $A \in \mathcal{B}$.
First, we must prove that $\mu$ is well-defined. It means, we must prove that if $A \subseteq \Bbb R$ is Borel measurable then $c(A\cap C)$ is Borel (or Lebesgue) measurable, so we can have $\rho(c(A \cap C))$.
Let $\Sigma =\{A \subseteq \Bbb R: \exists E \subseteq [0,1] \textrm{ Borel measurable, such that } A\cap C=c^{-1}(E) \}$. It is easy to prove that $\Sigma$ is a $\sigma$-algebra and that, since $c$ is monotone, every open interval is in $\Sigma$. So $\mathcal{B} \subseteq \Sigma$. So, if $A \subseteq \Bbb R$ is Borel measurable, there is $E$ Borel measurable such that $A\cap C=c^{-1}(E)$, so, since $c: C \rightarrow [0,1]$ is surjective, we have that $c(A\cap C)=E$ is Borel measurable. So $\mu$ is well-defined.
Let us now prove $\mu$ is a measure. We have $$\mu(\emptyset) = \rho(c(\emptyset \cap C)) =\rho(c(\emptyset )) = \rho(\emptyset) =0$$
Now, let $\Gamma =\{ y\in [0,1] : \exists a,b \in C, a\neq b \textrm{ and } y=c(a)=c(b)\} $. It is easy to prove that $\Gamma$ is countable and so $\rho(\Gamma)=0$.
Now let $A$ and $B$ be disjoint measurable subsets of $\Bbb R$. Then, $A \cap C$ and $B \cap C$ are disjoint measurable subsets of $C$. Note that $c(A \cap C)$ and $c(B \cap C)$ may not be disjoint, because $c$ (even restricted to $C$) is not injective, but $c(A \cap C) \cap c(B \cap C) \subseteq \Gamma$. So, $\rho(c(A \cap C) \cap c(B \cap C))=0$. So we have \begin{align*} \mu(A \cup B) & = \rho(c((A \cup B)\cap C)) =\\ & = \rho(c((A\cap C) \cup ( B \cap C))) = \\ &=\rho(c(A\cap C) \cup c(B\cap C)) = \\ & = \rho(c(A\cap C)) + \rho(c(B\cap C)) - \rho(c(A\cap C) \cap c(B\cap C)) = \\ & = \rho(c(A\cap C)) + \rho(c(B\cap C)) = \\ & = \mu(A) + \mu(B) \end{align*}
Now to prove that $\mu$ is $\sigma$-additive, it is enough to prove that $\mu$ is continuous from below.
Suppose $\{A_i\}_i$ is a non-decreasing monotonic family of measurable subsets of $\Bbb R$. Then $\{A_i \cap C\}_i$ is a non-decreasing monotonic family of measurable subsets of $C$. Then $\{c(A_i \cap C)\}_i$ is a non-decreasing monotonic family of measurable subsets of $[0,1]$. So, wee have
\begin{align*} \mu \left (\bigcup_i A_i \right )& = \rho \left (c \left ( \left ( \bigcup_i A_i \right )\cap C \right ) \right ) = \\ &= \rho \left ( \bigcup_i c(A_i \cap C) \right) = \\ &= \lim_i \rho(c(A_i \cap C)) = \\ & = \lim_i \mu(A_i) \end{align*}
So $\mu$ is $\sigma$-additive.
It is immediate that $\mu(\Bbb R)= \rho(c(\Bbb R \cap C))=\rho(c(C))=\rho([0,1])=1$. So $\mu$ is a probability. It is also immediate that if $A\cap C=\emptyset$ then $\mu(A)= \rho(c(A\cap C)) =\rho(c(\emptyset))=\rho(\emptyset) = 0$. So, a support of $\mu$ is $C$.