Are these probabilities orthogonal?

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Given $p\in (0,1)$ we define a Borel probability $\mu_p$ in the interval $[0,1]$.

We assign $\mu_p([0,1/2))=1-p$ and $\mu_p([1/2,1))=p$. We iterate this process by diving each interval in two and giving $(1-p)$ of the measure of the previous interval to the first side and $p$ of the measure of the previous interval to the second side. In this fashion we have $\mu_p([0,1/4))=(1-p)^2$, $\mu_p([1/4,1/2))=(1-p)p=\mu_p([1/2,3/4))$, $\mu_p([3/4,1))=p^2$, and so on and so forth. This defines a Borel measure because the dyadic intervals are a generator of the Borel $\sigma$-algebra. I'm asked whether these probabilitites are continuous ($\mu_p({x})=0\;\forall x\in[0,1)$) and whether, given $p\neq q$, $\mu_p\perp\mu_q$ (there's a Borel-measurable set $A$ such that $\mu_p(A)=0$ and $\mu_q(A^{\mathsf{c}})=0$).

I figured out the first part in the following way. Let $x\in[0,1)$. For every $n\in\mathbf{N}\;\exists\; 0\leq j\leq 2^n-1$ such that $x\in[j2^{-n},(j+1)2^{-n})$, and we have

\begin{equation} \mu_p({x})\leq\mu_p([j2^{-n},(j+1)2^{-n}))\leq max\{p^n, (1-p)^n\}\xrightarrow{\: n \to \infty \: }0 \end{equation}

Now, I'm really clueless for the second part. It seems like the probabilities shouldn't be orthogonal, as all intervals have positive probability for both probabilities and all singletons have probability $0$ for both probabilities. But how can we be sure that there isn't a strange collection of points concentrating all the measure for $p$ and no measure for $q$? Or, alternatively, how can we find such a collection?

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Here is another description of the measure $\mu_p$: Pick an iid sequence of bits $x_1,x_2,...$, with $P[x_i=1] = p$ and $P[x_i=0]=1-p$, and form the sum $x=\sum_{i=1}^\infty x_i 2^{-i}$. Then $x$ has probability measure $\mu_p$. (The interval $[0,1/2)$ has probability $1-p$ and the interval $[1/2,1]$ has probability $p$, corresponding to $x_1=0$ and $x_1=1$, etc.)

The strong law of large numbers tells us that $$P\left[\lim_{n\to\infty} \frac{x_1+\cdots+x_n}n = p\right] = 1.$$ You should be able to guess what the set $A$ is, and fill in the remaining details.