It is well-known that as you have more evidence (say in the form of larger $n$ for $n$ i.i.d. examples), the Bayesian prior gets "forgotten", and most of the inference is impacted by the evidence (or the likelihood).
It is easy to see it for various specific case (such as Bernoulli with Beta prior or other type of examples) - but is there a way to see it in the general case with $x_1,\ldots,x_n \sim p(x|\mu)$ and some prior $p(\mu)$?
EDIT: I am guessing it cannot be shown in the general case for any prior (for example, a point-mass prior would keep the posterior a point-mass). But perhaps there are certain conditions under which a prior is forgotten.
Here is the kind of "path" I am thinking about showing something like that:
Assume the parameter space is $\Theta$, and let $p(\theta)$ and $q(\theta)$ be two priors which place non-zero probability mass on all of $\Theta$. So, the two posterior calculations for each prior amount to:
$$p(\theta | x_1,\ldots,x_n) = \frac{\prod_i p(x_i | \theta) p(\theta)}{\int_{\theta} \prod_i p(x_i | \theta) p(\theta) d\theta}$$
and $$q(\theta | x_1,\ldots,x_n) = \frac{\prod_i p(x_i | \theta) q(\theta)}{\int_{\theta} \prod_i p(x_i | \theta) q(\theta) d\theta}$$
If you divide $p$ by $q$ (the posteriors), then you get:
$$p(\theta | x_1,\ldots,x_n)/q(\theta | x_1,\ldots,x_n) = \frac{p(\theta)\int_{\theta} \prod_i p(x_i | \theta) q(\theta)d \theta}{q(\theta)\int_{\theta} \prod_i p(x_i | \theta) p(\theta)d \theta}$$
Now I would like to explore the above term as $n$ goes to $\infty$. Ideally it would go to $1$ for a certain $\theta$ that "makes sense" or some other nice behavior, but I can't figure out how to show anything there.