# If $f_n: [0,T] \rightarrow \mathbb{R}$ are RCLL (i.e. càdlàg) with bounded $\sup_{n \in \mathbb{N}}f_n$, then $\sup_{n \in \mathbb{N}} f_n$ is RCLL

Assume we have a sequence $$(f_n)_{n \in \mathbb{N}}$$ of RCLL (i.e. càdlàg, i.e. right-continuous admitting left limits) functions $$f_n : [0,T] \rightarrow \mathbb{R}$$ with the property that, for every $$x \in [0,T]$$ there exists $$K > 0$$ with $$\sup_{n \in \mathbb{N}}|f_n(x)| \leq K$$. So we can define a function $$f^*(x):=\sup_{n \in \mathbb{N}}f_n(x)$$ for every $$x \in [0,T]$$ and see that $$f^*:[0,T] \rightarrow \mathbb{R}$$ .

My question now is, wheter we can say that $$f^*$$ is RCLL (i.e. càdlàg), since I cannot think of a counterexample, and cannot come up with a proof.

The functions $$f_n(x) = \begin{cases} 0 & x\leq0 \\ nx & x\leq1/n \\ 1 & x\geq1/n \end{cases}$$ are continuous, hence RCLL, but $$f(x) = \sup_{n} f_n(x) = \begin{cases} 0 & x\leq0 \\ 1 & x > 0 \end{cases}$$ is not right continuous.