Suppose we are given a function $f(x)$. We want to show the following claim: \begin{align} \liminf_{\epsilon \to \infty} \frac{f(x)}{f(x+\frac{t}{\epsilon} )}=1, \end{align} almost everywhere $(x,t)$ (in Lebesgue measure) .
Question: What is a minimal assumption on the points of discontinuity of $f$ that will guarantee this to be true?
For example, is it sufficient to assume that the set of points of discontinuity has Lebesgue measure zero? But then there are sets that are dense and have measure zero. I am somewhat confused about this.
Usually $\epsilon$ is a notation reserved for variables that tend to zero or are otherwise small, so I will use $r$ instead of $\epsilon$. For the limit \begin{align} \lim_{r \to \infty} \frac{f(x)}{f(x+\frac{t}{r} )} \tag{$\ast$} \end{align} to equal $1$ for a.e. $(x,t)$, It is necessary and sufficient that the set of discontinuity points of $f$ and the zero set $f^{-1}(0)$ both have Lebesgue measure zero. Indeed, if $f(x) \ne 0$ and $t>0$, then the limit $(*)$ equals $1$ if and only if $f$ is right-continuous at $x$, by the definition of right-continuity. Similarly, if $f(x) \ne 0$ and $t<0$, then the limit in $(\ast)$ equals $1$ if and only if $f$ is left-continuous at $x$.
The original question asked for $\liminf$ instead of $\lim$ in $(\ast)$ to equal $1$ for a.e. $(x,t)$. For this the zero set $f^{-1}(0)$ must still have Lebesgue measure zero, but the condition on the points of continuity is sufficient but no longer necessary.