Let $f:[0,\infty)\to\mathbb R$ be right-continuous and assume that the left-sided limit $$f(t-):=\lim_{s\to t-}f(s)$$ exists for all $t\ge0$. Now, let $$\Delta f(t):=f(t)-f(t-)\;\;\;\text{for }t\ge0,$$ $0\le T_0<T_1<T_1$ and $\varepsilon\ge0$.
How can we show that
- $\exists t\in(T_0,T_1]:|\Delta f(t)|>\varepsilon$
- $\exists k\in\mathbb N:\forall l\in\mathbb N:\exists s_1,s_2\in\mathbb Q\cap\left(T_0,T_1+\frac1l\right]:|s_1-s_2|<\frac1l\text{ and }|f(s_1)-f(s_2)|>\varepsilon+\frac1k$
are equivalent?
While intuitively reasonable, it's not clear to me how we need to construct the equivalence. For example, assume 1.: Since $f$ has a left-sided limit at $t$, we can minimize $|f(s)-f(t-)|$ arbitrarily by choosing $s<t$ sufficiently close to $t$. In the same way, by right-continuity at $t$, we can minimize $|f(s)-f(t)|$ arbitrarily by choosing $s\ge t$ sufficiently close to $t$. So, maybe the basic idea could be to choose $s_1<t$ and $s_2\ge t$ and use $$\varepsilon<|\Delta f(t)|\le|f(s_1)-f(s_2)|+|f(s_1)-f(t-)|+|f(s_2)-f(t)|\tag1.$$ (I've assumed $\varepsilon>0$. Guess the case $\varepsilon=0$ needs to be treated separately?)
As $|\Delta f(t)| > \epsilon$ we can choose $k \in \mathbb{N}$ such that $|\Delta f(t)| \geq \epsilon+3/k$. Since $f$ has a finite left-hand limit at $t$, we can choose $\delta_1>0$ such that
$$|f(u)-f(t-)| < \frac{1}{k} \quad \text{for all $u \in [t-\delta_1,t)$}.$$
Analogously, the right-continuity yields that there exists $\delta_2>0$ such that
$$|f(t)-f(v)| < \frac{1}{k} \quad \text{for all $v \in [t,t+\delta_2]$}.$$
If $\ell \in \mathbb{N}$ is a given fixed number we may assume without loss of generality that $\delta := \min\{\delta_1,\delta_2\} <1/(2\ell)$. Choose some $s_1 \in [t-\delta,t) \cap \mathbb{Q}$ and $s_2 \in [t,t+\delta) \cap \mathbb{Q}$. Then $$|s_1-s_2| \leq |s_1-t| + |t-s_2| \leq 2 \delta = \frac{1}{\ell}$$ and $$\begin{align*} \epsilon + \frac{3}{k} \leq |\Delta f(t)| &\leq |f(s_1)-f(t-)| + |f(t)-f(s_2)| + |f(s_1)-f(s_2)| \\ &< \frac{2}{k} + |f(s_1)-f(s_2)|, \end{align*}$$ i.e. $$|f(s_1)-f(s_2)| > \epsilon + \frac{1}{k} $$
Let $k \in \mathbb{N}$ be as in 2. and choose for any $\ell \in \mathbb{N}$ some numbers $s_1^{(\ell)}$ and $s_2^{(\ell)}$ in $[T_0,T_1+1]$ such that $$|s_1^{(\ell)}-s_2^{(\ell)}| \leq 1/\ell \tag{1}$$ and $$|f(s_1^{(\ell)})-f(s_2^{(\ell)})| > \epsilon+1/k. \tag{2}$$ Since the interval $[T_0,T_1+1]$ is compact we can find convergent subsequences, say $(t_1^{(\ell)})$ and $(t_2^{(\ell)})$, of $(s_1^{(\ell)})$ and $(s_2^{(\ell)})$, and it follows from $(1)$ that $$ t = \lim_{\ell \to \infty} t_1^{(\ell)} = \lim_{t \to \infty} t_2^{(\ell)}.$$ By the continuity assumptions on $f$, this implies
$$f_i := \lim_{\ell \to \infty} f(t_i^{(\ell)}) \in \{f(t),f(t-)\} \tag{3}$$
for $i=1,2$. On the other hand, we know from $(2)$ that
$$|f_1-f_2|> \epsilon + \frac{1}{k}>0, \tag{4}$$
and therefore $f_1 \neq f_2$. From $(3)$ we thus conclude that one of the limits equals $f(t)$ and th other limit equals $f(t-)$. Hence, by (4),
$$|\Delta f(t)|= |f(t)-f(t-)| > \epsilon.$$