A limit involving maximum of increments of a cadlag function. My textbook on stochastic processes states that if $f$ is cadlag then
 $$\max_{1\leqslant j \leqslant n2^m} |f(j/2^m)-f((j-1)/2^m)|\to \sup_{0<t\leqslant n} |f(t)-f(t-)|$$ for $m\to \infty$ and for all $n\geqslant 1$.
Could someone please help me realize why the limit is true.
As a bonus if someone could shed some light on why cadlag implies locally bounded, it would be much appreciated.
 A: I'll bet there are much neater proofs, but I haven't dealt with cadlag functions for a very long time, so the following proof may look a bit clumsy.
The statement is trivial if $f$ is continuous. So, let $S=\sup_{0<t\leq n} |f(t)-f(t-)|>0$. We first show that
\begin{equation}
\limsup\limits_m\max_{1\leq j \leq n2^m} |f(j/2^m)-f((j-1)/2^m)|\le S.\tag{1}
\end{equation}
Suppose the contrary that the limit superior is greater than $cS$ for some $c>1$. Then there exists a sequence of numbers $j_m$ such that $|f(j_m/2^m)-f((j_m-1)/2^m)|>cS$ when $m$ is large. Since $[0,n]$ is a compact interval, $\{j_m/2^m\}$ has a limit point $t\in[0,n]$. In other words, there exists two sequence of numbers $\{t_k\}$ and $\{t_k'\}$, with $t_k\ge t_k'$ for every $k$, such that $|f(t_k)-f(t_k')|>cS$ when $k$ is large and $\lim_{k\to\infty}t_k=\lim_{k\to\infty}t_k=t$. However, since $f$ is right-continuous, if $t_k'\ge t$ for an infinite subsequence of $\{t_k'\}$, we must have $|f(t_k)-f(t_k')|\to0$ for this subsequence, which is a contradiction because the difference should be greater than $cS$. Similarly, as $f$ has a left-limit, we cannot have $t_k<t$ for infinitely many $k$. Therefore we must have $t_k<t\le t_k$ when $k$ is large. But then $|f(t_k)-f(t_k')|\to|f(t)-f(t-)|\le S$, which is also a contradiction. Thus we conclude that $(1)$ must hold.
Next, we show that
\begin{equation}
\liminf\limits_m\max_{1\leq j \leq n2^m} |f(j/2^m)-f((j-1)/2^m)|\ge S-\epsilon\tag{2}
\end{equation}
for every $\epsilon>0$. Recall that if $f$ is a cadlag function defined on a compact interval $I$, then for every $\epsilon>0$, there are at most finitely many jumps of $f$ whose absolute sizes exceed $\epsilon$. Therefore, the supremum $S = \sup\limits_{0<\le t\le1}|f(t)-f(t-)|$ is actually attainable at finitely many points. Let $t\in(0,n]$ be one of those maximizers, i.e. $S=|f(t)-f(t-)|$. Now, for  every $m$, there exists a unique $j_m$ such that 
$(j_m-1)/2^m < t \le j_m/2^m$. Since $f$ is cadlag, this implies that for every $\epsilon>0$, we must have
$$\left\vert\left(f\left(\frac{j_m}{2^m}\right)-f(t)\right)-\left(f\left(\frac{j_m-1}{2^m}\right)-f(t-)\right)\right\vert<\epsilon$$
when $m$ is sufficiently large. Hence $|f(j_m/2^m)-f((j_m-1)/2^m)|\ge S-\epsilon$ when $m$ is large and $(2)$ holds. Now the statement in the textbook follows from $(1)$ and $(2)$.
Finally, every continuous function are locally bounded. Therefore every cadlag funtion $f$ is locally bounded on the right (because it is right-continuous) and locally bounded on the left (because if we redefine $f(x)$ as $\lim_{t\to x-}f(t)$, then $f$ is left-continuous at $x$; hence the redefined $f$ and the original $f$ are locally bounded around $x$ on the left).
