limit of sequence transformed random variable Let $f$ be some function with $\lim_{s\to\infty} f(s)=0$. Now consider the following inequality: $$f(s)<e^{-(s x)^2},$$ where $x$ is a random variable that is standard normal distributed. The question is whether the inequality is true for all possible realizations of $x \sim N(0,1)$ as $s\to \infty$. Clearly, both sides go to 0 in the limit. So whether or not this is true depends on the rate at which $f$ goes to 0 compared to the right hand side.
Now suppose we were to do a change of variables to $y=sx$ on the right hand side. Then an equivalent statement would be: Is $$f(s)<e^{-y^2}$$ true for all $y\sim N(0,\sigma^2)$ as $s\to\infty$? Now, however, fixing a particular $y$, the statement seems to be always true... 
I find this discrepancy somewhat odd and I suspect that something is wrong how I treat the transformation here. But I am confused about where exactly my reasoning goes wrong.
Here are some thoughts:
I know that the transformation is not well-defined in the limit, so this could be one problem. Another thing is whether one can even fix a particular $y$ and validate the inequality pointwise. E.g., it could be that generally, it is not sufficient to fix a particular $y$ and show that for any such $y$ there exists an $s$ such that the inequality is true. Alternatively, it could also be that just in this context one cannot keep $y$ fixed given that the transformation itself does depend on $s$ (or, equivalently, given that the distribution of $y$ does depend on $s$).
It would be great if someone could clarify where exactly my reasoning goes wrong.
 A: As you said, the answer seems like it depends specifically on the the function f(s). If you really want the answer to hold for every possible realization of the normal distribution however, and not just almost surely or in probability or something like that, then I would have to assume the answer is no for any function that did not go to zero extremely rapidly.
More precisely, N(0,1) can take any real value. So the inequality would literally have to fold for all x in R, which seems implausible to me in most cases.
If you use some function that goes to 0 faster than every conceivable exponential (maybe like the gamma function, since that generalizes the factorials on $R^+$?) it might have a chance of working. 
Otherwise if the function is $O(e^{as^2})$ for any $a\in R$, then it does not seem possible this could hold for every realization of N(0,1).
A: Just to put together what I wrote in the comments:
When you change the variables, the way to think about the second inequality would be in terms of $f(\frac{y}{x})$ and not f(s). I.e. one can not assume that in the second inequality s would be independent of y and still be deterministic.
If you fix y, and then let s go to infinity, you are as a result varying x, which then would no longer have the same N(0,1) distribution, i.e. fixing y is a different problem, since even though s and x are independent, s and y are not. 
In other words, if you are fixing y, then s is actually a random variable and not deterministic, i.e. $\frac{y}{x}$, taking values anywhere on ℝ and Cauchy distributed (see https://en.wikipedia.org/wiki/Ratio_distribution#Gaussian_ratio_distribution), for which the probability that |s| is very large (and hence f(s) smaller than any arbitrary number) is 1−ϵ, since the Cauchy distribution has "very thick tails", e.g. enough so that its expectation does not even exist. 
Thus the second inequality should be true with very high probability, although it does not mean what you want it to mean. 
If you are fixing s, then s is measurable with respect to the trivial sigma algebra (since it is a constant RV), so (y|s) (y given s) would also be measurable with respect to the trivial sigma algebra, so essentially any event corresponding to (y|s), which seems to be your second inequality, is either a.s. true or a.s. false, which corroborates the inequality being a.s. true. 
If in the second inequality you are taking s to be fixed, then the random variable you are considering is actually (y|s), which has to be measurable with respect to a trivial sigma algebra (since s is a constant RV), so hence (y|s) would also be a.s. constant ($s\mathbb{E}x = 0$), so the inequality would become f(s) < 1 as s goes to infinity, which is obviously true by assumption.
Changing variables for random variables always requires great care, since RVs are actually real-valued functions from a probability space, not numbers, so the logic involved will always be more complicated and the reasoning used for regular variables unsuited.
EDIT: to address concerns about the notation used in this answer, and to appeal to those who are familiar with the concept of conditional expectation as a random variable, (y|s) is not standard notation, but it is obviously highly suggestive of (and meant to denote) $\mathbb{E}(Y|\sigma(s))$, which is a constant random variable (it is a constant RV because of the conditioning on the trivial sigma algebra, the sigma is being trivial because the RV s is a constant RV).
