$\dim V_\lambda=\dim V_{w(\lambda)}$ when $V$ is an integrable module over a Kac-Moody algebra 
Let $V$ be an integrable module over a Kac-Moody algebra. Then $\dim V_\lambda=\dim V_{w(\lambda)}$ for each $\lambda\in\mathfrak{h}^*$ and $w\in W$ (the Weyl group).

It's a proposition stated in Kac's book Infinite dimensional Lie algebras, however he just states it follows form the two following lemmas:


*

*$\textrm{ad}\ e_i$ and $\textrm{ad}\ f_i$ are locally nilpotent over a Kac-Moody algebra.





*If $\lambda$ is a weight of $V$, then $\lambda-\lambda(\alpha_i^\vee)\alpha_i$ is also a weight of the same multiplicity.


However I don't see why this is the case?
Also the first lemma is pretty clear to me, but I don't see why the second lemma holds?
I'm also happy to see, if there is another way of proving this. Maybe it can somehow be reduced to the finite-dimensional case like when showing that "The module $L(\Lambda)=M(\Lambda)/N(\Lambda)$ (the quotient of a Verma module) over a Kac-Moody algebra is integrable if and only if $\lambda(\alpha_i^\vee)\geq 0$ for all $i=1,\cdots,n$"?
It should also be noted, that it've been earlier shown that "If $\alpha\in\Delta, w\in W$ then $w\alpha\in W$ and $\textrm{mult}(\alpha)=\textrm{mult}(w\alpha)$."
Here $\Delta$ denotes the union of the positive and negative roots.

I've managed to prove that $V$ decomposes into a direct sum of finite-dimensional irreducible $\mathfrak{sl}_2$-modules.
Now in order to finish the proof, we need to justify that given a module $N$ in the $\mathfrak{sl_2}$-decomposition then $\dim N_\lambda=\dim N_{r_i\lambda}$. I think this i exactly the second lemma stated by Kac above. However we also need to justify that the weights have the form Joppy mentions. We also need to justify that $\lambda(\alpha_i^\vee)\in\mathbb{Z}$. This is where I'm still in doubt.
 A: When $\mathfrak{g}$ is a Kac-Moody algebra with Cartan indexing set $I$, the vector space $\mathfrak{g}_{(i)} := \mathbb{C}\{e_i, \alpha_i^\vee, f_i\}$ is a Lie subalgebra isomorphic to $\mathfrak{sl}_2$ for each $i \in I$. An $\mathfrak{h}$-diagonalisable $\mathfrak{g}$-module $V$ is integrable if it is a locally finite $\mathfrak{g}_{(i)}$-module for all $i \in I$. (Recall that to be locally finite means that $V$ is a union of finite-dimensional modules). This means that an integrable module $V$ decomposes as a (possibly infinite) direct sum of finite-dimensional $\mathfrak{g}_{(i)}$-modules, or in other words a direct sum of $\mathfrak{sl}_2$-modules.
The weights of finite-dimensional $\mathfrak{sl}_2$-modules are symmetric around zero, with the reflection $s: \lambda \mapsto - \lambda$ giving the symmetry. Similarly, the weights of $\mathfrak{h}$-invariant finite-dimensional $\mathfrak{g}_{(i)}$-modules are symmetric around the hyperplane $\ker \langle \alpha_i^\vee, - \rangle \subseteq \mathfrak{h}^*$, with the simple reflection $r_i(\lambda) = \lambda - \langle \alpha_i^\vee, \lambda \rangle \alpha$ giving the symmetry. If we were to zoom in on a single finite-dimensional $\mathfrak{g}_{(i)}$-submodule in $V$, with highest weight vector $v_{\Lambda}$ say, then we would find that its weights were
$$ \Lambda, \Lambda - \alpha, \Lambda - 2\alpha, \cdots, r_i(\Lambda) + \alpha, r_i(\Lambda). $$
What I picture in my head is that I have the large representation $V$, and when I only look at a single $i \in I$, it breaks down into finite $\mathfrak{sl}_2$-strings. On this finite-dimensional $\mathfrak{g}_{(i)}$-submodule $U \subseteq V$ we have $\dim U_\lambda = \dim U_{r_i \lambda}$ for all $\lambda$, therefore adding this up across all the submodules we get $\dim V_\lambda = \dim V_{r_i \lambda}$. Since we can do this for any $i \in I$, we get $\dim V_\lambda = \dim V_{w \lambda}$ for any element $w$ of the Weyl group.
