Prove that $\nu$ is absolutely continuous w.r.t. $\mu$ iff $\sum \alpha_j^2<\infty$ This is question 3 in Chapter 4.12 from Barry Simon - Real analysis

I tried to define $ f_j : \{0,1\} \to \mathbb{R} $ where $n$ means that it is a function from $j$th $\{0,1\}$ in the product to $\mathbb{R}$. It is given by
$$ f_j(0)=2p_j \ \ \ f_j(1)=2(1-p_j)$$
Finally, let $f=\otimes f_j$ which gives me, by Kolmogorov Consistency Theorem, $d\nu=fd\mu$ (???). But, I believe this is wrong or I am missing something because I didn't even use the assumption.
Idea behind defining $f_j$ in this way was, consider $\{0,1\}^N$ then for example for the point $(1,1,...1)$ the integral gives ($f^N=\otimes_{j=1}^N f_j$)
$$ \int_{\{(1,1,...,1)\}}f^N(x)d\mu(x)=\prod_{j=1}^N \frac{1}{2}2(1-p_j)=\int_{\{(1,1,...,1)\}}d\nu(x) $$
Then I tried to find counter example but I couldn't even find a counter example for the case $\alpha_j=1/4$ for all $j$.

*

*Maybe we can use the fact $X=\{0,1\}^\infty$ is Cantor set in product topology?


*$d\nu_p$ should be $d\mu_p$ obviously.
Any hint and solution is appreciated.
 A: I don't know how much of a hint this is, but this is an application of a result  due to Kakutani (Kakutani's dichotomy).  There are several books on probability that present this result based on the theory of Martingales, for example Durrett, R. Probability: Theory and Examples, 5th edition, pp. 229-230.
I hope other solutions present an analysis point of view.

Following the assumptions in the OP, let
$\nu_{p}=p\delta_0+(1-p)\delta_1$
Set $\mu=\nu_{1/2}$. For any $p\in(0,1)$ define
$$f_p(\omega):=\frac{d\nu_p}{d\mu}(\omega)=2p\mathbb{1}_{\{0\}}(\omega)+2(1-p)\mathbb{1}_{\{1\}}(\omega)$$
Kakutani's dichotomy's theorem states that for $F(d\omega)=\prod^\infty_{k=1}\nu_{p_k}(d\omega_k)$ and $G=\prod^\infty_{k=1}\mu(d\omega_k)$, then either $F\ll G$ or $F\perp G$ according to whether $\prod_n\int\sqrt{f_{p_n}(\omega_n)}\,\mu(d\omega_n)>0$ or $\prod_n\int\sqrt{f_{p_n}(\omega_n)}\,\mu(d\omega_n)=0$.
Notice that
$$\int \sqrt{f_{p_k}(\omega)}\,\mu(d\omega)=\frac{p^{1/2}_k+(1-p_k)^{1/2}}{2^{1/2}}$$
In the setting of the OP, $p_k=\frac12+\alpha_k$.
\begin{align}\prod_n\int\sqrt{f_{p_n}(\omega_n)}\,\mu(d\omega_n)&=\prod_n\Big(\frac{p^{1/2}_n+(1-p_n)^{1/2}}{2^{1/2}}\Big)=\prod_n\Big(\frac{(1/2+\alpha_n)^{1/2} +(1/2-\alpha_n)^{1/2}}{2^{1/2}}\Big)\\
&=\prod_n\Big(\frac{(1+2\alpha_n)^{1/2}+(1-2\alpha_n)^{1/2}}{2}\Big)\\
&=\prod_n\Big(1+\frac{(1+2\alpha_n)^{1/2}+(1-2\alpha_n)^{1/2}-2}{2}\Big)
\end{align}
Converges of the product (to nonzero limit) occurs iff
$$\sum_n\Big|\frac{(1+2\alpha_n)^{1/2}+(1-2\alpha_n)^{1/2}-2}{2}\Big|<\infty$$
From
$$(1+x)^{1/2}=\sum^\infty_{n=0}\binom{1/2}{n}x^n$$
we have that
$$(1+x)^{1/2}+(1-x)^{1/2}=2\sum^\infty_{n=0}\binom{\alpha}{2n}x^{2n}=2-\frac14x^2-\frac{10}{128}x^4-\ldots$$
Thus
$$\frac{|(1+2\alpha_n)^{1/2}+(1-2\alpha_n)^{1/2}-2|}{2}=
\frac12\alpha^2_k+o(\alpha^2_k)$$
Alternatively
$$\begin{align}
\frac{(1+x)^{1/2}+(1-x)^{1/2}-2}{x^2}&\sim\frac{(1-x)^{1/2}-(1+x)^{1/2}}{4x(1-x^2)^{1/2}}\\
&=-\frac{1}{2(1-x^2)^{1/2}\big((1+x)^{1/2}+(1-x)^{1/2}\big)}\xrightarrow{x\rightarrow0}-\frac14
\end{align}$$
Thus
$$\frac{|(1+2\alpha_n)^{1/2}+(1-2\alpha_n)^{1/2}-2|}{2}\sim\frac{\alpha^2_k}{2}$$
