Let $f,g:X\rightarrow Y$ be morphisms from a locally Noetherian scheme $X$ to a separated scheme $Y$ that agree on an open dense subset $U\subseteq X$, and suppose that $U$ contains all the associated points of $X$. How can one show that $f=g$?
We know that $f$ and $g$ agree on a closed subscheme of $X$ that contains $U$ so it is enough to prove that every closed subscheme of $X$ containing $U$ is $X$, but how can we deduce this?