I am currently trying to understand how consistency of the likelihood ratio estimator follows from Kakutani's theorem:
Kakutani's theorem
Let $(X_n)_{n\ge1}$ be a sequence of independent non-negative random variables of mean $1$.
Set $a_n:=E(\sqrt{X_n})\in(0,1]$ and define $M_0:=1,\,M_{n\ge 1}:=\prod_{i=1}^{n}X_i$. Then $(M_n)_{n\ge 0}$ is a non-negative martingale and $M_n\to M_{\infty}$ a.s. for some random variable $M_\infty$. Moreover,
(a) if $\prod_n a_n>0,$ then $M_n\to M_{\infty}$ in $L^1$ and $E(M_{\infty})=1,$
(b) if $\prod_n a_n=0,$ then $M_{\infty}=0$ a.s.
Corollary
Let $P$ and $\tilde{P}$ be probability measures on a measurable space $(\Omega, \mathcal{F})$. Let $(X_n)_{n\ge 1}$ be a sequence of random variables. Assume that, under $P$ (respectively $\tilde{P}$), the random variables $X_n$ are independent and $X_n$ has law $\mu_n$ (respectively $\tilde{\mu}_n$) for all $n$.
Suppose that $\tilde{\mu}_n=f_n\mu_n$ for all $n$.
Define the likelihood ratio $L_n:=\prod_{i=1}^{n}f_n(X_n)$. Then, under $P$,
(a) if $\prod_n \int_{\mathbb{R}}\sqrt{f_n}d\mu_n>0$, then $L_n$ converges a.s. and in $L^1$,
(b) if $\prod_n \int_{\mathbb{R}}\sqrt{f_n}d\mu_n=0$, then $L_n\to 0$ a.s.
In particular, if $\mu_n=\mu$ and $\tilde{\mu}_n=\tilde{\mu}$ for all $n$ with $\mu\neq \tilde{\mu}$, then $P(L_n\to 0)=1,\ \tilde{P}(L_n\to\infty)=1$.
Since $f_n(X_n)$ is a random variable with mean $1$ under $P$, I can understand that $P(L_n\to 0)=1$ in the corollary. However, where does $\tilde{P}(L_n\to\infty)=1$ come from?
Furthermore, I have read that this implies consistency of the likelihood ratio statistic $L_n$ but for this we would have to fix real parameters for $\mu$ and $\tilde{\mu}$. From the corollary, the obvious and only possible choices for $\mu$ and $\tilde{\mu}$ would be $0$ and $\infty$, respectively. The problem here is that $\infty\not\in\mathbb{R}$, or am I missing something e.g. that $\mathbb{R}\cup\{\infty\}$ is allowed as parameter?