Problem with infinite product measures Given some measurable space $\left(X,\mathcal{F}\right)$ and two probability measures $\mu$ and $\nu$ on this space one can define
$$H_{\theta}(\mu,\nu)=\int\left(\frac{d\mu}{d\lambda}\right)^{\theta}\left(\frac{d\nu}{d\lambda}\right)^{1-\theta}d\lambda
 $$ 
where $\theta\in[0,1]
 $ and $\lambda$ is any positive measure such that $\mu,\nu<<\lambda$ (it is in fact independent of the choice of this measure, and we can in fact assume it is also a probability measure).
Now let $\mathbb{X}$ be a polish space, $\mathcal{F}=\mathbb{B}(\mathbb{X})
 $, and consider the countable product of measurable spaces $\left(\prod_{n}\mathbb{X},\otimes_{n}\mathcal{F}\right)
 $
, and the following measures $
\mu=\prod_{n}\mu_{n},\:\nu=\prod_{n}\nu_{n}
$, on this space. Where $\mu_{n}$ and $\nu_{n}$ are probability measures on $\left(\mathbb{X},\mathcal{F}\right)
 $. 
I wish to show that $H_{\theta}(\mu,\nu)=\prod_{n=1}^{\infty}H_{\theta}(\mu_{n},\nu_{n}).$ 
It is trivial for the case $\theta=0,1$, so assume $\theta\in(0,1)$.
I can show that $H_{\theta}(\mu_{1}\times\mu_{2},\nu_{1}\times\nu_{2})=H_{\theta}(\mu_{1},\nu_{1})H_{\theta}(\mu_{2},\nu_{2})$
From which it follows that, $$H_{\theta}(\mu,\nu)=\prod_{n=1}^{N}H_{\theta}(\mu_{n},\nu_{n})H_{\theta}(\prod_{n=N+1}^{\infty}\mu_{n},\prod_{n=N+1}^{\infty}\nu_{n}).
 $$
And since (by Hölder's inequality), $$H_{\theta}(\prod_{n=N+1}^{\infty}\mu_{n},\prod_{n=N+1}^{\infty}\nu_{n})\leq1,$$
it only remains to show that $$H_{\theta}(\prod_{n=N+1}^{\infty}\mu_{n},\prod_{n=N+1}^{\infty}\nu_{n})\geq1.
 $$
Since then $$H_{\theta}(\mu,\nu)=\lim_{N}\prod_{n=1}^{N}H_{\theta}(\mu_{n},\nu_{n})\lim_{N}H_{\theta}(\prod_{n=N+1}^{\infty}\mu_{n},\prod_{n=N+1}^{\infty}\nu_{n})=\prod_{n=1}^{\infty}H_{\theta}(\mu_{n},\nu_{n}).
 $$
I am kind of stuck here. Maybe this idea is going nowhere. I am not very familiar with infinite product measures and their integrals. Any pointers would be appreciated.
 A: In general,
$$H_{\theta} \left( \prod_{n=N+1}^{\infty} \mu_n, \prod_{n=N+1}^{\infty} \nu_n \right)=1$$
does not hold. Simply consider $X=\mathbb{R}$ endowed with the Lebesgue measure $\lambda$ and set $d\mu_n(x) := 1_{[0,1]}(x) \, d\lambda(x)$ and $d\nu_n(x) := 1_{[2,3]}(x) \, d\lambda(x)$. Then, obviously, $$H_{\theta}(\mu,\nu) = H_{\theta} \left( \prod_{n=N+1}^{\infty} \mu_n, \prod_{n=N+1}^{\infty} \nu_n \right)= H_{\theta}(\mu_n,\nu_n)=0$$ since the densities have disjoint support.

So here is an alternative way to prove the equality of the measures:
Let
$$p_n := \frac{d\mu_n}{d\lambda} \qquad \qquad q_n := \frac{d\nu_n}{d\lambda}.$$
Then we see from the definition of $\mu$ that
$$p := \frac{d\mu}{d\lambda} = \prod_{n=1}^{\infty} p_n. \tag{1}$$
Similarly,
$$q := \frac{d\nu}{d\lambda} = \prod_{n=1}^{\infty} q_n.$$
In order to prove $H_{\theta}(\mu,\nu) = \prod_{n=1}^{\infty} H_{\theta}(\mu_n,\nu_n)$, it suffices to show that
$$H_{\theta}(\mu,\nu)(B) = \prod_{n=1}^{\infty} H_{\theta}(\mu_n,\nu_n)(B), \qquad B \in \mathcal{D}$$
for a $\cap$-stable generator $\mathcal{D}$ of the product $\sigma$-algebra $\prod_{n} \mathcal{F}$. Such a generator is given by the cylinder sets, i.e.
$$\mathcal{D} := \{ B; B=\left(\prod_{j=1}^n X\right) \times B_{n+1} \times \ldots \times B_{n+k} \times \prod_{j \geq 1} X \, \text{for some} \, B_{n+1},\ldots,B_{n+k} \in \mathcal{B}(X).\}$$
Now, if $B$ is of such a form, it is not difficult to see from the definition of the product measure and $(1)$ that the equality holds.
A: We have the following counterexample. Let $\mu_k$ and $\lambda_k$ be Gaussian linear measures on $R$ with parameters $(0,1)$ and $(1,1)$.  Then  $\prod_{k \in  N}\mu_k \perp \prod_{k \in  N}\lambda_k$ because
(i) $\prod_{k \in  N}\mu_k \{ (x_k)_{k \in N}: \lim_{n \to \infty}\frac{\sum_{k=1}^nx_k}{n}=0\}=1$,
(ii) $\prod_{k \in  N}\lambda_k \{ (x_k)_{k \in N}: \lim_{n \to \infty}\frac{\sum_{k=1}^nx_k}{n}=1\}=1$,
and
(iii) $\{ (x_k)_{k \in N}: \lim_{n \to \infty}\frac{\sum_{k=1}^nx_k}{n}=0\} \cap \{ (x_k)_{k \in N}: \lim_{n \to \infty}\frac{\sum_{k=1}^nx_k}{n}=1\}=\emptyset.$
The validity of conditions (i)-(ii) follows from the strong law of large numbers(Hint: under random variables we can take  usual projections). The validity of (iii) is obvious.
Now setting $\nu_k=\lambda_k$ for $k \in N$, we get 
$H_1(\mu,\nu)=\int_{R^{\infty}}(\frac{d\mu}{d \lambda})^1(\frac{d\nu}{d \lambda})^0 d \lambda=0 $(because $(\frac{d\mu}{d \lambda})^1=0$) and $\prod_{k \in N}H_1(\mu_k,\nu_k)=\prod_{k \in N}\int_R((\frac{d\mu_k}{d \lambda_k})^1(\frac{d\nu_k}{d \lambda_k})^0d \lambda_k =\prod_{k \in N}\int_R d\mu_k =1$.
Hence $$H_1(\mu,\nu)\neq \prod_{k \in N}H_1(\mu_k,\nu_k).$$
