Is there a simple elegant ways to show that the uniform limit of a sequence of cadlag functions on a compact set (say of the real line) is also cadlag? I seem to mess up in dealing with moving discontinuity points and my proof becomes way too involved for what should be a simple fact.

I am aware that we cannot remove the request that the limit be uniform. But, is the result true if we removed the request that the functions are defined on compact sets?

Thank you. Maurice


2 Answers 2


Where the limit $f$ is continuous there is nothing to prove so let's consider only the points at which $f$ is discontinuous. Let $x$ be a point where $f$ is discontinuous.

(I): $\lim_{t\to x^-}f(t)$ exists

For eacn $n\in \mathbb{N}$, let $f_n^-(x)=\lim_{t\to x^-}f_n(t)$, whose existence is assured because the sequence is cadlag.

Lemma: $f_n^-(x)$ is Cauchy.

Proof: Let $\epsilon >0$.

Since $f_n\to f$ uniformly, there is some $n_0 \in \mathbb{N}$ such that $n \geq n_0$ implies $|f_n(t)-f(t)| \leq \epsilon/6$. It follows that if $m,n \geq n_0$ then $|f_n(t) - f_m(t)| \leq \epsilon/3$.

Now, given $m,n \geq n_0$, from $f_m^-(x)=\lim_{t\to x^-}f_m(t)$ and $f_n^-(x)=\lim_{t\to x^-}f_n(t)$ we obtain that there are $\delta_m,\delta_n >0$ such that

$$t \in (x-\delta_m,x) \implies |f_m(t)-f_m^-(x)| < \epsilon/3\\ t \in (x-\delta_n,x) \implies |f_n(t)-f_n^-(x)| < \epsilon/3$$

In particular, if $\delta=\min\{\delta_m,\delta_n\}>0$ and $t \in (x-\delta,x)$, then both inequalities are satisfied simultaneously. In this case, if $m,n\geq n_0$ then:

\begin{align} |f_n^-(x)-f_m^-(x)|\leq \underbrace{|f_n^-(x)-f_n(t)|}_{<\epsilon/3} +\underbrace{|f_n(t)-f_m(t)|}_{<\epsilon/3} +\underbrace{|f_m(t)-f_m^-(x)|}_{<\epsilon/3}<\epsilon, \end{align}

which proves the lemma. $\square$

By the lemma, and the completeness of $\mathbb{R}$, $\lim_{n\to\infty}f_n^-(x)=f^-(x)$ exists.

Claim: $\lim_{t\to x^-}f(t)=f^-(x)$

Indeed, we have that

$$|f(t)-f^-(x)|\leq \underbrace{|f(t)-f_n(t)|}_{(1)} +\underbrace{|f_n(t)-f_n^-(x)|}_{(2)} +\underbrace{|f_n^-(x)-f^-(x)|}_{(3)} $$

$(1)$ goes to $0$ as $n \to \infty$ from the uniform convergence of $f_n\to f$.
$(2)$ goes to $0$ as $t\to x^-$ from the limit $f_n^-(x)=\lim_{t\to x^-}f_n(t)$.
$(3)$ goes to $0$ as $n\to\infty$ from the limit $\lim_{n\to\infty}f_n^-(x)=f^-(x)$.

Together, these mean we can make $|f(t)-f^-(x)|$ as small as we want by taking $t<x$ close enough to $x$. In other words, this proves the claim, and consequently also part (I). $\square$

(II): $\lim_{t\to x^+}f(t)$ exists and equals $f(x)$

We can basically use the proofs of part (I) here.

For eacn $n\in \mathbb{N}$, let $f_n^+(x)=\lim_{t\to x^+}f_n(t)$. Since the sequence is cadlag, these limits all exist and moreover $f_n^+(x)=f_n(x)$.

A proof similar to the lemma will show that $f_n^+(x)$ is Cauchy. But of course it is! $f_n\to f$ uniformly, so $f_n^+(x)=f_n(x)$ converges pointwise to $f(x)$. As a convergent sequence, it is Cauchy!

Let $f^+(x)=\lim_{n\to\infty}f_n^+(x)$. Well, we know from the previous paragraph that $f^+(x)=f(x)$.

A proof similar to the claim will show that $\lim_{t\to x^+}f(t)=f^+(x)$. Because we know that $f^+(x)=f(x)$, we get that $\lim_{t\to x^+}f(t)=f(x)$, which is exactly what we wanted to prove for part (II). Therefore, we are done! $\square$

Notice that we did not need the assumption of compactnes.

  • $\begingroup$ Thank you for the detailed answer. I do appreciate your help. $\endgroup$
    – Maurice
    Feb 12, 2017 at 6:56

Suppose $f_n\to f$ uniformly on $D\subset R$, where $f_n$ are cadlag.

For right-continuity of $f$ at $t∈ D$, take $n$ with $|f_n-f|<\epsilon$ uniformly, and $a>0$ with $|f_n(t)-f_n(s)|<\epsilon$ for $s\in[t,t+a)∩ D$. By triangle inequality, we get $|f(t)-f(s)|<3\epsilon$ for $s\in[t,t+a)∩ D$.

For left-hand limits, suppose that $t_k\uparrow t$. Then, with $|f_n-f|<\epsilon$ uniformly, $$ f_n(t-)-\epsilon = \liminf_k f_n(t_k)-\epsilon ≤ \liminf_k f(t_k) ≤ \limsup_k f(t_k) ≤ \limsup_k f_n(t_k)+\epsilon = f_n(t-)+\epsilon, $$ which shows that $f(t_k)$ must have a limit $f(t-)$, since $\epsilon>0$ is arbitrary. Suppose that $f(t-)$ is not unique: $s_k\uparrow t$ and $t_k\uparrow t$ are such that $\lim_k f(s_k)$ and $\lim_k f(t_k)$ differ by $3\epsilon$. But then for $|f_n-f|<\epsilon$ uniformly, limits $\lim_k f_n(s_k)$ and $\lim_k f_n(t_k)$ differ at least by $\epsilon$, a contradiction.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .