Absolute continuity of measures on infinite binary sequences Suppose $P$ and $Q$ are two probability measures on the space $\Omega = \{0,1\}^{\mathbb N}$ of infinite binary sequences equipped with the product $\sigma$-algebra generated by its cylinder sets, with
$$ p_k = P(\omega_k = 1) \in (0,1), \quad q_k = Q(\omega_k = 1) \in [0,1].$$
Question: Why does $Q\ll P$ hold if and only if 
$$\sum_k \left( 1 - \sqrt{p_k q_k} - \sqrt{(1-p_k)(1-q_k)}   \right) < \infty?$$
A special case has already appeared on MO. I found this question through social media, where people were sharing a test from the Indian Statistical Institute that featured Donald Trump. A scan of the test is here. The question I'm asking is 4(a), and part (b) follows from it after some algebra. I was able to solve the other questions on the exam but could not make much progress on this.
 A: This is a particular case of Kakutani's theorem on equivalence of product measures. There is a very detailed exposition in Chapter III, Section 9 of Shiryaev's Probability. 
A: Ok,  Loïc Teyssier, Gerry Myerson, Nik Weaver, Stefan Waldmann, coudy, and a few other people will scold me badly for answering a question about a relatively simple exercise in undergraduate probability instead of crying loudly that "We do research here!" and closing, but here goes.
Note first of all that all examiners have an irritating habit of write the conditions in a slick form that hides what is going on as much as possible. So your first task on any exam is to trade beauty for clarity.
Since $2\sqrt(xy)=x+y-(\sqrt x-\sqrt y)^2$ and $(\sqrt x-\sqrt y)^2\asymp\frac{(x-y)^2}{x+y}$, the condition really is
$$
\sum_k (p_k-q_k)^2\left[\frac 1{(p_k+q_k)}+\frac{1}{(1-p_k)+(1-q_k)}\right]<+\infty
$$
Let $\Delta_k=q_k-p_k$.
Now the formal density of $Q$ with respect to $P$ is 
$$
\prod_k \left(1+\frac{\Delta_k}{p_k}\right)^{\omega_k}\left(1-\frac{\Delta_k}{1-p_k}\right)^{1-\omega_k}
$$
We want to show that the partial products $D_n=\prod_{k=1}^n(\dots)$ converge in $L^1(P)$. The trivial sufficient condition would be the uniform $L^2(P)$ bound and we almost have it. Indeed, 
$$
\mathcal E_P D_n^2=\prod_{k=1}^n\left[1+\Delta_k^2\left(\frac 1{p_k}+\frac{1}{1-p_k}\right)\right]
$$
So everything would work if we had the series without $q$'s in the denominator convergent. Unfortunately, we do not. Note, however, that if  $\frac{\Delta_k^2}{p_k}>10\frac{\Delta_k^2}{p_k+q_k}$, then the quantity in the right is at least $\frac 12q_k$, so the sum over such $k$ of $q_k$ converges and, thereby, with $Q$-probability $1$ only finitely many of such $\omega_k$ are $1$. Similarly for $1-$ fractions. Thus, if we just condition upon the corresponding finite sets of bad indices (and finite subsets of integers are countably many, so we just partition the interesting part of the probability space into countably many chunks), we can use the $L^2$ criterion in each chunk separately.
This is sufficiency. Now, since it is an undergraduate exercise, do the other part yourself (or ask the people mentioned above to help you). I'll stop here.
