Part (1) follows from Morphisms, Lemma \ref{morphisms-lemma-birational-isomorphism-over-dense-open}. Part (2) follows from part (3) and Lemma \ref{lemma-finite-in-codim-1} (and the fact that finite morphisms have finite fibres). \medskip\noindent Part (3) follows from part (4) and Morphisms, Lemma \ref{morphisms-lemma-finite-fibre} but we will also give a direct proof. Let $x \in X$ be as in (3). By Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood} we may assume $f$ is finite. We may assume $X$ affine. This reduces us to the case of a finite birational morphism of Noetherian affine schemes $Y \to X$ and $x \in X$ such that $\mathcal{O}_{X, x}$ is a normal domain. Since $\mathcal{O}_{X, x}$ is a domain and $X$ is Noetherian, we may replace $X$ by an affine open of $x$ which is integral. Then, since $Y \to X$ is birational and $Y$ is reduced we see that $Y$ is integral. Writing $X = \Spec(A)$ and $Y = \Spec(B)$ we see that $A \subset B$ is a finite inclusion of domains having the same field of fractions. If $\mathfrak p \subset A$ is the prime corresponding to $x$, then $A_\mathfrak p$ being normal implies that $A_\mathfrak p \subset B_\mathfrak p$ is an equality. Since $B$ is a finite $A$-module, we see there exists an $a \in A$, $a \not \in \mathfrak p$ such that $A_a \to B_a$ is an isomorphism. \medskip\noindent Let $x \in X$ and $y \in Y$ be as in (4). After replacing $X$ by an affine open neighbourhood we may assume $X = \Spec(A)$ and $A \subset \mathcal{O}_{X, x}$, see Properties, Lemma \ref{properties-lemma-ring-affine-open-injective-local-ring}. Then $A$ is a domain and hence $X$ is integral. Since $f$ is birational and $Y$ is reduced it follows that $Y$ is integral too. Consider the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$. This is a ring map which is essentially of finite type, the residue field extension is finite, and $\dim(\mathcal{O}_{Y, y}/\mathfrak m_x\mathcal{O}_{Y, y}) = 0$ (to see this trace through the definitions of quasi-finite maps in Morphisms, Definition \ref{morphisms-definition-quasi-finite} and Algebra, Definition \ref{algebra-definition-quasi-finite}). By Algebra, Lemma \ref{algebra-lemma-essentially-finite-type-fibre-dim-zero} $\mathcal{O}_{Y, y}$ is the localization of a finite $\mathcal{O}_{X, x}$-algebra $B$. Of course we may replace $B$ by the image of $B$ in $\mathcal{O}_{Y, y}$ and assume that $B$ is a domain with the same fraction field as $\mathcal{O}_{Y, y}$. Then $\mathcal{O}_{X, x} \subset B$ have the same fraction field as $f$ is birational. Since $\mathcal{O}_{X, x}$ is normal, we conclude that $\mathcal{O}_{X, x} = B$ (because finite implies integral), in particular, we see that $\mathcal{O}_{X, x} = \mathcal{O}_{Y, y}$. By Morphisms, Lemma \ref{morphisms-lemma-morphism-defined-local-ring} after shrinking $X$ we may assume there is a section $X \to Y$ of $f$ mapping $x$ to $y$ and inducing the given isomorphism on local rings. Since $X \to Y$ is closed (by Schemes, Lemma \ref{schemes-lemma-section-immersion}) necessarily maps the generic point of $X$ to the generic point of $Y$ it follows that the image of $X \to Y$ is $Y$. Then $Y = X$ and we've proved what we wanted to show.