Ergodicity of the action of $\operatorname{SL}(n,\mathbb R)$ on $\operatorname{SL}(n,\mathbb R)/\operatorname{SL}(n,\mathbb Z)$ $\DeclareMathOperator\SL{SL}$Let $G\mathrel{:=}\SL(n,\mathbb R)$ and $\Gamma\mathrel{:=}\SL(n,\mathbb Z)$. Consider the action of $G$ on $(G/\Gamma,\mu)$ by left translation, where $\mu$ is the Borel probability measure on the homogeneous space $G/\Gamma$ that is left-invariant w.r.t $G$. I wonder where I can find the proof that the action of $G$ is ergodic?
If this is true in more general settings say when $G$ is a simple Lie group as in the $G=\SL(n,\mathbb R)$ case, please let me know.

Many comments below mentioned the Howe-Moore theorem. But I am aware of the fact (as a corollary of Howe-Moore) that every unbounded subgroup of $G$ also acts ergodically/mixingly after we proved that $G$ itself acts ergodically.
 A: The question is why a measurable function which is invariant under the action of all $g \in G$ must be a.e. constant. But since the action of $G$ is transitive on $G/\Gamma$ it is easy to see that this is the case: take a non constant function $\varphi$ on $G/\Gamma$, there are points $x,y$ such that in small neighborhood $U$ and $V$ of each where the function takes different values in at least 99% of the measure of each. Now, taking some $g$ that maps $x$ to $y$ you can see that $\varphi$ cannot be $g$-invariant.
Once this is seen it is that one can apply Howe-Moore to deduce that every unbounded subgroup also acts ergodically (in fact, mixing). Improvements exist as Asaf points out.
A: Posting YCor's comment to hopefully end some confusion -
of course details are needed. But it seems that Fubini applied to $|f(gx)-f(x)|$ on $G\times X$ implies that for a.e. all $x$ we have $f(gx)=f(x)$ for a.e. all $g$. Since for given $x$ and every measure-generic subset $U$ of $G$ we have $Ux$ measure-generic in $X$ (for $G$ second-countable and $X$ being $G$ mod discrete this seems quite clear), it follows that $f$ is a.e. constant.
A: This is equivalent to for an irrational element $g, g \in G / \Gamma$, irrational mean the equivalent class $[g]\cap \Gamma=\emptyset$,
$$
\lim _{N \rightarrow+\infty} \sum_{n=1}^{N} \delta_{g^ n} \longrightarrow \text { Haar Messure } \quad (*)
$$
And $(*)$ can be proved by calculate the paring $\sum_{n=1}^{N} \delta_{g^ n}$ with the periodic function on $\mathrm{SL}(n, \mathbb{R})$ induced by the lattice $\mathrm{SL}(n, \mathbb{Z})$ in it.
I do not know what is the explicit expression of the periodic function, say them are $\phi$.
But I believe its behavior is the same as $e(nx)=e^{2\pi inx}$ in the case of $\mathbb{T}$, and in particular use the Gauss summation method on $\phi$ will give us the desirable bound for indicating Weyl criterion is true.

I will give the detail in the above argument later, and this argument is stronger than Howe-Moore Theorem because it can give the speed of convergence to mixing.
