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Let $f:[0,infty)tomathbb R$ be right-continuous and assume that the left-sided limit $$f(t-):=lim_{sto t-}f(s)$$ exists for all $tge0$. Now, let $$Delta f(t):=f(t)-f(t-);;;text{for }tge0,$$ $0le T_0<T_1<T_1$ and $varepsilonge0$.

How can we show that

1. $exists tin(T_0,T_1]:|Delta f(t)|>varepsilon$


2. $exists kinmathbb N:forall linmathbb N:exists s_1,s_2inmathbb Qcapleft(T_0,T_1+frac1lright]:|s_1-s_2|<frac1ltext{ and }|f(s_1)-f(s_2)|>varepsilon+frac1k$

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 $sge t$ sufficiently close to $t$. So, maybe the basic idea could be to choose $s_1<t$ and $s_2ge t$ 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?)

1. implies 2.:

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 $delta := min{delta_1,delta_2} <1/(2ell)$. 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. $$|f(s_1)-f(s_2)| > epsilon + frac{1}{k} $$

2. implies 1.

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),

$$|Delta f(t)|= |f(t)-f(t-)| > epsilon.$$