Let $f$ be a rational function on a variety $V$. Let $U = \{P\in V; f \textrm{ is defined at }P\}$. Then $f$ defines a function from $U$ to $k$. Show that this function determines $f$ uniquely. So a rational function may be considered as a type of function, but only on the complement of an algebraic subset of V, not on V itself.
I thought as follows: if there is an other $g\in k(V)$ such that $g(P)=f(P)$ for all $P\in U\cap \widehat{U}$ where $\widehat{U}=\{Q\in V;g \textrm{ is defined at }P\}$ then need conclude that $f=g$ in $k(V)$. I am sure?