Almost everywhere convergence of a function of rv Let f(x) a continuous function on $\mathbb {\bar{R}^+}$(extended positive real line,  $x\in(0,\infty]$). Take $y\in \mathbb R^+$. We can say that $\lim_{y\to 0}\frac{x}{y}=\infty $  almost everywhere because the set on which $\lim_{y\to 0}\frac{x}{y}\ne\infty $ is a subset of $\mathbb Q$ which would have a Lebesgue measure zero. Does this make sense? is there a better way to say the same things?
Also, because $\lim_{y\to 0}\frac{x}{y}=\infty $  a.e. and $f$ is continuous on $\mathbb {\bar{R}}$ we can write $f(\lim_{y\to 0}\frac{x}{y})=\lim_{y\to 0}f(\frac{x}{y})$. Does this make sense as well?
Now, if instead of $x$ a real number, we have a random variable X. Can we say that $f(\lim_{y\to 0} \frac{X}{y})=\lim_{y\to 0 }f(\frac{X}{y})$ a.e., knowing that $f(\infty)=c$? If so, the motivation should involve a probability measure, right?
Any reference or hint is appreciated.
Thank you!
 A: I'm going to answer your questions in order below. Note this is partially opinion-based so take these answers with a grain of salt.


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*I'd prefer to see you explicitly state the subset of $\mathbb{Q}$ you're referring to (the one for which $\lim_{y\to0}\frac{x}{y}\neq\infty$). In fact, you don't really need to point out that it's a subset of $\mathbb{Q}$ to conclude it has measure $0$, just that it is countable (or, in this case, a singleton). Otherwise I don't have a problem with your phraseology.

*Yes, $f(\lim_{y\to0}\frac{x}{y}) = \lim_{y\to0}f(\frac{x}{y})$ makes sense here, so long as you add a.e. after the equation.

*While probability theory/real analysis isn't my field, I'd say that this makes sense (or at least can be made precise). The only issues I have with it are that (a) I think you meant to write $\lim_{y\to0}$ not $\lim_y\to\infty$, (b) you should specify which type of random variable you're referring to, and (c) it's more common to say that the value of this limit is almost surely $\infty$, not $\infty$ a.e. (This is just special terminology used in probability theory.) Hopefully someone else will come along who can elaborate on this point or disagree with me!
