Can I embed $\mathbb R^{\mathbb N}$ with a partial order into $^\ast\mathbb{R}$ with the linear order?

Question

Define a relation $\prec$ on $\mathbb R^{\mathbb N}$ as,

For all $f, g \in \mathbb R^{\mathbb N} $, $f \prec g$, if for all $n \in \mathbb N$, $f(n) \leq g(n)$, and there exists a $m \in \mathbb N$ such that $f(m) < g(m)$.

Does there exist a function $u : \mathbb R^{\mathbb N} \to {^\ast\mathbb{R}}$ such that

For all $f, g \in \mathbb R^{\mathbb N} $,$f \prec g$ implies that $u(f) < u(g)$ ?

${^\ast\mathbb{R}}$ denotes hyperreal numbers. I tagged economics, because this is an attempted extension of utility representation theorem.

Answer 1

If I have understood the construction of ${}^\ast\Bbb R$ properly, then the function defined as follows should work.

$$u(f)_n=\sum_{k=0}^nf(k)$$

Here $u(f)_n$ denotes the $n$-th member of the hyperreal $u(f)$. If $f\prec g$, then $u(f)_n<u(g)_n$ for all $n\ge m$. There are cofinitely many $n$ greater than $m$, so regardless of the choice of ultrafilter, $u(f)<u(g)$.