Proving that the Bernoulli self similar measure is doubling 
Let $\mu_p$ a measure which is the push forward of the bernouli product measure $(p,1-p)^\mathbb N$. Let S=$\{f_1,\dots f_m\}$ an IFS, a system of functions with attractor $K$, means $$K=\bigcup_{i=1}^nf_n(K).$$
  a. Prove that for $K=[0,1],S=\{\frac{x}{2},\frac{x+1}{2}\}$ and for any $p\neq \frac 1 2$, the measure $\mu_p$ is not doubling its support. 
b. Now consider the same $K$ but $S=\{\frac{x}{3},\frac{x+1}{3},\frac{x+2}{3}\}$. The mass is now divided according to probability vector $(p,1-p-q,q)$. Prove tat $\mu_p$ is doubling if and only if $p=q$

I don't really know from where to start. I understand that each singleton on this interval gets weight according to the probability $p,q$ or $1-p-q$ but I can't connect it to terms of balls and doubling measures. Can I get any hint for both parts?
 A: The general topic you are studying is called self-similar measures.  Such a measure can be approximated by a sequence of simpler measures in much the way that a self-similar set can be approximated by a sequence of simpler sets.  Let's examine how the self-similar measure $\mu$ with probability list $\{2/3,1/3\}$ is generated by the IFS 
$$S=\{f_1(x),f_2(x)\}=\{x/2,(x+1)/2\}.$$
The invariant set of $S$ is the unit interval, which will also be the support of $\mu$. As a zeroth approximation $\mu_0$ to $\mu$, we'll choose Lebesgue measure restricted to the unit interval.  The sequence of measures will be defined recursively.  Given a Borel set $A$, $\mu_{n+1}(A)$ will be defined in terms of $\mu_{n}$ by
$$\mu_{n+1}(A) = \frac{2}{3}\mu_n(f_1^{-1}(A)) + \frac{1}{3}\mu_n(f_2^{-1}(A)).$$
For example, $\mu_1([0,1/2])$ satisfies
\begin{align}
\mu_{1}([0,1/2]) &= \frac{2}{3}\mu_0(f_1^{-1}([0,1/2])) + \frac{1}{3}\mu_0(f_2^{-1}([0,1/2])) \\
 &= \frac{2}{3}\mu_0([0,1]) + \frac{1}{3}\mu_0([1,2]) \\ 
 &= \frac{2}{3}\cdot1+\frac{1}{3}\cdot0=\frac{2}{3}.
\end{align}
The other half, similarly satisfies $\mu_1([1/2,1])=1/3$.  Thus, application of the IFS has the effect of redistributing the measure in a sense.  We can visualize the process using a sequence of histograms.  Here's what Lebesgue measure looks like together with the first five steps in the process.

And, here's a higher order approximation:

Of course, there is a limit measure (in the weak-$*$ topology) that satisfies
$$\mu(A) = \frac{2}{3}\mu(f_1^{-1}(A)) + \frac{1}{3}\mu(f_2^{-1}(A))$$
for all Borel sets $A$.
Now, look at what is happening near $1/2$.  The measure just to the right of $1/2$ is much heavier than the measure just to the left of $1/2$.  Specifically, they can be determined as the terms in the binomial expansion of
$$\left(\frac{2}{3}+\frac{1}{3}\right)^n.$$
I think this should give you more than enough to do part (a). Just imagine applying the doubling criterion at a point like $2^{-1}-2^{-n}$.

Part (b) should rely on very similar ideas.  You should understand that these ideas generalize quite a lot.  Thus, you can have IFS with $m$ functions $\{f_1,f_2,\ldots,f_m\}$ and you can define a recursive process like
$$\mu_{n+1}(A) = \sum_{i=1}^m p_i \, \mu_n(f_i^{-1}(A))$$
to get a measure $\mu$ satisfying 
$$\mu(A) = \sum_{i=1}^m p_i \, \mu(f_i^{-1}(A)).$$
You should think about how the measure accumulates on either side of the $1/3$ or $2/3$, perhaps using a trinomial expansion.
