Birational invariance of cohomology Let $X$ be a smooth projective scheme over an algebraically closed field $k$. I have heard it claimed that $\dim_k H^i(X, \mathcal{O}_X)$ is a birational invariant of $X$ for all $i$. Does anyone know where to find a proof of this?
 A: Karl's nice answer settles the question in general. Let me give here a different proof in the case where $X$ is smooth over $\mathbb{C}$, which involves Dolbeault theorem instead of Leray spectral sequence.
Assume that $X$ is smooth over $\mathbb{C}$ and let $f \colon X \dashrightarrow Y$ be a birational map. By Dolbeault isomorphism, it is sufficient to prove that 
$H^0(X, \Omega_X^i)=H^0(Y, \Omega_Y^i) \quad (*)$ 
for all $i$. Let $\omega \in H^0(Y, \Omega_Y^i)$ be a holomorphic $i$-form on $Y$. Then $f^* \omega$ is a meromorphic $i$-form on $X$, which is holomorphic outside the indeterminacy locus $Z \subset X$ of $f$. Since the codimension of $Z$ is at least $2$ and the poles of a meromorphic form are a divisor, it follows that $f^* \omega$ is actually holomorphic on the whole of $X$. This enables us to define an injective map
$f^* \colon H^0(X, \Omega^i_Y) \to H^0(X, \Omega_X^i)$.
Since $f$ is birational it follows that $f^*$ has an inverse, so we have proven $(*)$. 
EDIT. After writing this answer, I realized that it essentially coincides with the one given by Dan Petersen for the question quoted in the comments. Anyway, I think there is no harm in leaving it here. 
