Edited version of question:

Suppose $f:\mathbb{R}\longrightarrow \mathbb{R}$ is cadlag (that is, it is right continuous and its left limits exist) and $g:\mathbb{R}\longrightarrow \mathbb{R}$ is continuous. Then is it necessarily true that $g\circ f:\mathbb{R} \longrightarrow \mathbb{R}$ is also cadlag?

I am trying to prove this more general statement for fun, but I have been quite stuck for a while now.

PS: Currently for what I want to prove all I really need is the weaker statement that $g$ to be the mapping $x\mapsto x^2$. However I am curious whether this more general statement holds as I am sure it will become handy later on/it is in general a useful fact to know.

Many thanks to spaceisdarkgreen for pointing out my horrendous mistake!

  • $\begingroup$ It is easy to show that $g\circ f$ is right continuous, right? because both $f,g$ are right continuous. So you only have to show that the limit from the left exists. $\endgroup$
    – Yanko
    Jan 3, 2018 at 19:39
  • $\begingroup$ Yes I gathered that =) But that is precisely where I am getting stuck, showing the left limit exists... $\endgroup$
    – User086688
    Jan 3, 2018 at 19:41
  • $\begingroup$ You mean $f\circ g.$ $\endgroup$ Jan 3, 2018 at 19:56
  • $\begingroup$ Or rather either you mean $f\circ g : \mathbb R\to A$ and it's false (and its falsity has nothing to do with whether $A=\mathbb R$... it's just that if $g$ is decreasing the direction of approach is reversed) or you mean $g\circ f: \mathbb R\to \mathbb R$ and it's true cause $g$ preserves limits. $\endgroup$ Jan 3, 2018 at 20:04
  • $\begingroup$ Actually the latter only makes any sense for $A=\mathbb R,$ so I'm not sure why wrote it down, guess I was still confused. $\endgroup$ Jan 3, 2018 at 20:17

1 Answer 1


For existence of left limits, since $g$ is continuous, we have $$\lim_{x\to a^-} g(f(x)) =g(\lim_{x\to a^-}f(x))$$ and the RHS exists since $f$ is cadlag.

Similarly, for right continuity we have $$\lim_{x\to a^+}g(f(x)) = g(\lim_{x\to a^+}f(x)) = g(f(a)).$$


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