TITLE: Natural extension of a discontinuous function
QUESTION [5 upvotes]: Let $u : \mathbb{R} \to \mathbb{R}$ be the right continuous version of the Heaviside step function. What does the natural extension $u^*$ of $u$ to the set $\mathbb{R}^*$ of the hyperreals look like? Specifically, what does it look like on an infinitely narrow symmetric interval around $0$?
This question arose out of the example I used in Why it is absolutely mistaken to cancel out differentials?. Specifically, if 
$$u_n(x)=\begin{cases} 0 & x \in (-\infty,-1) \\ (x+1)^n & x \in [-1,0] \\ 1 & x \in (0,\infty) \end{cases}$$ 
then $u_n \to u$ pointwise but (as far as I can tell) $u_n^* \not \to u^*$ pointwise.

REPLY [3 votes]: As you note, every real function $f$ has a natural hyperreal extension $f^\ast$.  This is true regardless of whether $f$ is continuous or not. Consider for example the Heaviside function $h$.  This is defined by $h(x)=0$ if $x\leq 0$ and $h(x)=1$ if $x>0$. By the transfer principle, the same conditions will be satisfied for $h^\ast$ when $x$ is hyperreal.  Thus for all negative or zero hyperreal $x$ one will have $h(x)=0$ whereas for all positive hyperreal $x$ one will have $h(x)=1$. In particular at all negative infinitesimals $h$ will be zero, whereas at all positive infinitesimals $h$ will be equal to one.