How to prove that càdlàg (RCLL) functions on $[0,1]$ are bounded?

Question

While studying the space $\mathbb{D}[0,1]$ of right continuous functions with left hand limits (i.e. càdlàg functions) on $[0,1]$, I came across the following theorem:

Theorem. If $f$ is càdlàg on $[0,1]$, it is bounded.

My proof attempt: I am aware that if a function has both left and right hand limits on $[0,1]$, then the set of discontinuities is at most countable. Hence I tackled this in two parts, one where the discontinuities are finite, the other infinite.

I got the finite discontinuity one. But I am stuck at the infinite discontinuity part. My guess is that there is something special about the countable discontinuities (e.g. they cannot be dense in $[0,1]$) and somewhere, I'll have to use the sequential compactness property to get a contradiction, but I am unable to collect my ideas.

I request any starting hints on this. A sketch of the proof would also be appreciated.

Answer 1

If $f$ is unbounded, there will be an infinite sequence of points $(x_n)_{n\in\mathbb{N}}$ such that $\lim_{n\to\infty}x_n=x\in[0,1]$ and $\lim_{n\to\infty}f(x_n)=\infty$. Split the subsequence into two, one on the left and one on the right of $x$. One of them must be an infinite sequence. Suppose both of them are infinite (the other cases are similar). Call them $(u_n)_{n\in\mathbb{N}}$ and $(v_n)_{n\in\mathbb{N}}$ respectively. Then $u_n\to x^-,\ v_n\to x^+$ and $\lim_{n\to\infty}f(u_n)=\lim_{n\to\infty}f(v_n)=\infty$, but this is impossible because càdlàg functions have left and right limits everywhere.

Answer 2

Thanks for your help everyone. I think I got it. Kindly check my proof for errors if any:

Proof: $\forall x \in [0,1]$, fix $\epsilon > 0$. Now because at every point, both left and right hand limits exist(except at $0$ and $1$, in which case only one of the following will be there),

$$\exists \delta_x,\eta_x > 0$$ $$x_0 \in (x-\delta_x,x) \Rightarrow |f(x_0) - f(x^-)| < \epsilon$$ $$x_0 \in (x,x+\eta_x) \Rightarrow |f(x_0) - f(x)| < \epsilon$$

where $f(x^-)$ is the left hand limit of the function at x and since we have right continuity, $f(x^+) = f(x)$.

Let $O_x = (x-\delta_x,x)$ and $V_x = (x,x+\eta_x)$. Let $W_x = O_x\cup V_x \cup\{x\}$, which is still an open set. Then $$\bigcup_{x \in [0,1]} W_x$$ is an open cover for $[0,1]$. Since $[0,1]$ is compact, it has a finite subcover. Hence $$ [0,1] \subseteq \bigcup_{k=1}^n W_{x_k}$$

For $k=1...n$, if $x \in O_{x_k}$, then $|f(x)-f(x^-_k)| < \epsilon \Rightarrow |f(x)| < |f(x^-_k)| + \epsilon$

If $M_1 = \max_{k=1...n} |f(x_k^-)|$, then $|f(x)| < M_1 + \epsilon$. (Only if $x$ belongs to any of the "$O$" balls)

Similarly, by defining $M_2 = \max_{k=n+1...m} |f(x_k)|$, and following similar steps, we get $|f(x)| < M_2 + \epsilon$. (For x belonging to "$V$" balls).

Finally, if $x \in \{x_1,\cdots, x_n\}$, then $|f(x)| \le M_3 = \max_{1\le i\le n} |f(x_i)|$.

So the bound for $x \in [0,1]$ is, $$|f(x)| < \max(M_1,M_2,M_3) + \epsilon$$ Thus f is bounded. $\blacksquare$

I am interested in other elegant proofs if there are any. Thanks again. Update: Thanks to $\epsilon-\delta$ for the comment. I have fixed the proof, I think. Please check it.

Answer 3

Billigsley gives an excellent one-line proof. I am repeating it here. First note that this Lemma is true :

For every $\epsilon >0$, $ \exists$ partition $ 0=t_0<t_1<\ldots,t_k=1$ such that for the set $S_i = [t_i,t_{i+1})$, we have $\sup_{s,t \in S_i} |f(s)-f(t)| <\epsilon$ for all $i$.

This Lemma is easily proved. Once Lemma is seen to be true, choose an $\epsilon >0$. Then the bound on $f$ is simply $(\sum_{l=1}^{k} J_{l} ) + \epsilon \times k$ (first term is the amount of jumps which occur at the $k$ points which were obtained using the Lemma, while the second term is the increment of $f$ within these intervals.