I have been trying to understand the following proof but I don't fully understand the implicit last steps. From the replacement of the groups to the deducing of the theorems. I wondered if anyone could help me out explaining it to me in an intuitive manner.
Let $G= SLd(R) \times \mathrm{SL}_d(\mathbb{R})$, $\Gamma=\mathrm{SL}_d(\mathbb{Z}) \times \mathrm{SL}_d(\mathbb{Z})$, $H_0=\left\{(g, g) \mid g \in \mathrm{SL}_d(\mathbb{R}\}\right.$, $\widetilde{A}=\left\{\left(a, a^{-1}\right) \mid a \in A\right\}$ for $A$ positive diagonal subgroup in $SLd(R)$ and $\widetilde{N}=\left\{\left(n_1 ,n_2^{\mathrm{t}}\right) \mid n_1, n_2 \in N\right\} $ for $N$ upper triangular unipotent group in $SLd(R)$.
I have already proved that $P=\tilde{N} \widetilde{A}H_0$ contains the identity in its interior.
$g H_0 \widetilde{g} \Gamma$ equidistributes as $g H_0 \rightarrow \infty$ in $G / H_0$
PROOF. By Lemma 12.15 we may safely assume that $g=\widetilde{a} \in \widetilde{A}^{+}$. Let $\kappa>0$ be arbitrarily small, and choose $K \subseteq H_0 \widetilde{g} \Gamma$ compact such that $$ m_{H_0 \tilde{g} \Gamma}(K)>(1-\kappa) m_{H_0 \tilde{g} \Gamma}\left(H_0 \widetilde{g} \Gamma\right) . $$
Fix $f \in C_c(X)$ and $\varepsilon>0$, and choose $\delta>0$ smaller than the injectivity radius on $K$ and small enough to ensure that $$ \mathrm{d}\left(x_1, x_2\right)<\delta \Longrightarrow\left|f\left(x_1\right)-f\left(x_2\right)\right|<\varepsilon $$
Set $P=\tilde{N} \widetilde{A}$ and let $B=B_\delta^P$ be the $\delta$-neighborhood of $I \in P$. Now replace $H_0 \widetilde{g} \Gamma$ first by $K$ and then by $B K$, use the mixing property (Theorem 2.7), use the fact that $g_n=\widetilde{a_n} \in \widetilde{A}^{+}$contracts $N$, and deduce the proof of the theorem.