Understand the proof that rational resolution is independent of the resolution EDIT. Karl Schwede has given a different approach to solve this question, which is very clear. However, I still want to figure out how to deal with this problem along Ein's line, and waiting for the answer about my QUSETION A. B. below.
This is a duplication  of a question from MSE.
I am currently reading the notes Lectures on singularities and adjoint linear systems written by Ein and encounter some difficulties as follows, see Prop1.12.
Definition (Rational Singularity). A morphism $f: Y \rightarrow X$ is said to be a rational resolution if $Y$ is smooth and $f$ is a proper and birational morphism such that $R^{i} f_{*} \mathscr{O}_{Y}=0$ for $i>0$.
Proposition. Let $f: Y \rightarrow X$ be a rational resolution and $f^{\prime}: Y^{\prime} \rightarrow X$ be another resolution. Then $f^{\prime}: Y^{\prime} \rightarrow X$ is also a rational resolution.
Proof. We have a birational map $\varphi: Y -\to Y^{\prime}$. Successively blowing up the undefined locus of $\varphi$, we get a variety $Z$ and two proper birational morphisms $g: Z \rightarrow Y$ and $g^{\prime}: Z \rightarrow Y^{\prime}$ such that $h:=f \circ g=f^{\prime} \circ g^{\prime}$. Since $g$ is the composition of blowing-ups. Then $R^{q} g_{*}\left(\mathscr{O}_{Z}\right)=0$ for $q>0$. Apply the Leray spectral sequence
$$
E_{2}^{p, q}=R^{p} f_{*}\left(R^{q} g_{*}(\mathscr{F})\right) \Rightarrow R^{p+q}(f \circ g)_{*}(\mathscr{F}) .
$$
It follows that $R^{i} h_{*} \mathscr{O}_{Z}=0$ for $i>0$. Apply the Leray spectral sequence to $f^{\prime} \circ g^{\prime}$. It is easy to see that $R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$. In fact, it fits in the following exact sequence
$$
0 \rightarrow R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime} \rightarrow R^{1} h_{*} \mathscr{O}_{Z} \rightarrow f_{*}^{\prime} R^{1} g^{\prime} \mathscr{O}_{Z}
$$
Since $Y^{\prime}$ is smooth hence $Y^{\prime}$ has a rational resolution. Now $Z$ is another resolution of $Y^{\prime}$. By the above argument, we can conclude that
$R^{1} g_{*}^{\prime} \mathscr{O}_{Z}=0$. Apply the Leray spectral sequence to $p+q=2$. We see that $R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$. Hence $R^{2} g_{*}^{\prime} \mathscr{O}_{Z}=0$. By induction, we conclude that $R^{p} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$ for $p>0$.

Here is my question:
QUESTION A. Suppose we know that $R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$ and $R^{1} g_{*}^{\prime} \mathscr{O}_{Z}=0,$ how to deduce $R^{2} g_{*}^{\prime} \mathscr{O}_{Z}=0$ by applying the Leray spectral sequence as Ein said rather than using the five-term sequence, that is
$$
0 \rightarrow R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime} \rightarrow R^{1} h_{*} \mathscr{O}_{Z} \rightarrow f_{*}^{\prime} R^{1} g_*^{\prime} \mathscr{O}_{Z}\rightarrow R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}\rightarrow R^{2} h_{*} \mathscr{O}_{Z}?
$$
We can say $R^{2} f_{*}^{\prime} \mathscr{O}_{Z}=0$  since its prior one and posterior one both vanish.
I wonder how to use Leray spectral sequence at $p+q=2$, since now, we have
$$0=R^{2} h_{*} \mathscr{O}_{Z}=R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}\oplus R^{1} f_{*}^{\prime}(R^1 g_{*}^{\prime} \mathscr{O}_{Z})\oplus R^{2} g_{*}^{\prime} \mathscr{O}_{Z}$$
with the middle term vanishing, but we cannot say anything about the vanishing of $R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}$ or $R^{2} g_{*}^{\prime} \mathscr{O}_{Z}$.
QUESTION B. How to do the induction if we cannot use the five-term sequence for larger $p$, $q$?
 A: This is not exactly what is asked for, rather it is a different proof of the fact in question.
I assume you are working in characteristic zero with varieties.  I'm also going to assume you want $X$ to be normal (for simplicity, your rational resolutions will all factor through the normalization).  The proof in other characteristics is hard I think (see relevant papers by Kovács and A. Chatzistamatiou and K. Rülling).  The way I prefer to see this is via Grothendieck duality and also utilizing Grauert–Riemenschneider vanishing.  If you want to work with Q-Schemes with dualizing complexes, everything below works by the version of Grauert–Riemenschneider vanishing by Murayama (but let's stick with varieties).
Ok, if $X$ has a rational resolution $f : Y \to X$, then your hypothesis means that
$$
O_X \to R f_* O_Y
$$
is an isomorphism in the derived category.  In particular, by Grothendieck duality, you have that
$$
R f_* \omega_Y^{\bullet} \to \omega_X^{\bullet}
$$
is an isomorphism in the derived category as well.  But $Y$ is smooth and so Cohen-Macaulay, and in particular $\omega_Y^{\bullet} = \omega_Y[d]$ (ie, it's a shifted sheaf).  Furthermore by Grauert–Riemenschneider vanishing we have that  $R\pi_* \omega_Y[d] = \pi_* \omega_Y[d]$ is a sheaf.  Thus $\omega_X^{\bullet}$ is a sheaf too and $X$ is Cohen–Macaulay.  In particular, we now have that

*

*$X$ is Cohen-Macaulay.

*$\pi_* \omega_Y = \omega_X$.

Ok, now if $f' : Y' \to X$ is another resolution, we can find $Y''$ birationally mapping to both $Y$ and $Y'$.  Furthermore we may assume $Y'$ a smooth variety (you can resolve indeterminacies and then resolve the singularities, or take the product $Y \times_X Y'$, take the irreducible component dominating $X$ and resolve that).
In particular, since we need to show that $O_X \to R \pi_* O_{Y'}$ is a quasi-isomorphism, by Grothendieck duality it suffices to show that $g_* \omega_{Y'} = \omega_X$.  To do this, it suffices to show that both
$\pi_* \omega_{Y''} = \omega_{Y}$ and $\nu_* \omega_{Y''} = \omega_{Y'}$ (where $\pi$ and $\nu$ are the relevant maps) since then $f_* \pi_* \omega_{Y''} = \omega_X$ and chasing the diagram the other way we get $g_* \omega_{Y'} = g_* \nu_* \omega_{Y''} = \omega_X$.
Ok, so in particular we need to show that any resolution of a smooth variety is a rational resolution.  But to do that, it is a computation (that I can show some details if you want) that $\pi_* \omega_{Y''} = \pi_* \pi^* \omega_Y \otimes O_Y(D)$ where $D$ is an effective divisor.  Ie, the relative canonical over a smooth variety is effective (you might have seen this in other places in notes of Ein).  It follows now that $\pi_* \omega_{Y''} = \omega_Y$ which is what we wanted to show.
