Extension of Likelihood-density Given an observation scheme, say $(X_1,X_2,\ldots,X_n)$ where $X_i$ evolves to one measure, one can define the likelihood function, a parameter dependend density of the form
$$
L_n:=L(X_1,\ldots,X_n\mid\theta)
$$
This is a density wrt to a  measure say $P^n$. Thus there exists a measure e $Q^n$, such that $L_n$ is the radon-nikodym-derivative. However, for assumptions about consistency of an estimator one lets $n\rightarrow \infty$ can we expect a measure $Q^{\infty}?$
 A: Edit: I misinterpreted "consistency of an estimator" to be referring to Kolmogorov's extension theorem, as opposed to statistical consistency of an estimator, so what follows below is not right.
Usually the likelihood function factors into $L_n:=L(X_1\mid\theta)\cdots L(X_n\mid\theta)$. 
In this case the associated measure of likelihood function on $Q^\infty$ will exist by virtue of the Kolmogorov extension theorem, so long as your finite measure satisfies consistency conditions listed in the link.
A: Everything in the answer by "Alex R." is wrong.
You have a family of probability measures parametrized by $\theta$ in some parameter space $\Theta.$
You have a joint density function $(x_1,\ldots,x_n) \mapsto f_\theta(x_1,\ldots,x_n)$ whose domain is a Cartesian power of the space in which $X_1$ lies.
You have a likelihood function $\theta\mapsto L_n(\theta) = f_\theta(x_1,\ldots,x_n).$
This is a function of $\theta$ with $x_1,\ldots,x_n$ fixed at the actually observed data values.
You have a measure $\pi(d\theta)$ of some kind on the parameter space $\Theta.$ Often this is the prior probability distribution of $\theta.$
Then the likelihood function $L_n(\theta)$ is the Radon–Nikodym derivative with respect to $\pi$ of a measure on the parameter space, given by
$$
A \mapsto \int_A L_n(\theta) \pi(d\theta).
$$
If $\pi$ is the prior probability distribution of $\theta,$ then if one normalizes this measure by multiplying by a constant, then one gets the posterior probability distribution of $\theta{:}$
$$
\Pr(\theta\in A\mid X_1=x_1,\ldots,X_n=x_n) = \frac{\int_A L_n(\theta) \pi(d\theta)}{\int_\Theta L_n(\theta) \pi(d\theta)}.
$$
If $X_1,X_2,X_3,\ldots\mid\theta\sim\operatorname{i.i.d.}$ then (perhaps under some regularity assumptions?) the posterior probability measure will converge in probability to $\theta.$
