This is lemma 2.1.8, pg 19, Andreas Grathman, Algebraic Geometry
def 1: A regular function in $p\in X$, is an element of the following ring: $$\mathcal O_{X,\,p}=\{\frac{f}{g}\;:\;f,g\in k[X_1,\ldots,X_n]/I(X),\;g(p)\neq 0\}$$ Moreover a regular function on $U\subseteq X $ open is an element of the ring $$\mathcal O_X(U)=\bigcap_{p\in U} \mathcal O_{X,\,p}$$
The following defintiion of regular function is equivalent to the one of ring of regular functions.
def 2: Let $U$ be an open susbset of an affine variety $X \subseteq \Bbb A^n$. A set theoretic map $\varphi:U \rightarrow k$ is called regular at the point $P \in U$, if there is a neighborhood $V$ of $P$ in $U$ such that there are polynomials $f,g \in k[x_1, \ldots, x_n]$ with $g(Q) \not= 0$, and $\varphi(Q) = \frac{f(Q)}{g(Q)}$ for all $Q \in V$. It is called regular on $U$ if it is regular at every point in $U$.
The proof has a step which I could not follow.
So for each $\varphi$ regular in def 2, for a $P \in U$, we assign $f/g \in K(X)$ as given in def 2. So it suffices to prove $f/g \in O_X(U)$.
The author wrote, we may
...suppose that there are polynomials $f'/g'$ such that $f/g=f'/g'$ on some neighborhood $V'$ of $P'$.
why can we suppose this? What if there are no such poylnomials?
I believe I may be interpreting this completely wrong. May someone please elaborate my mis understanding?