Are Integrable $\mathfrak g$-modules and locally finite-dimensional $\mathfrak g$-modules equivalent definitions? Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra of rank $n$.  Set $I= \{1, \cdots, n\}$ and fix a set of Chevalley generators $\{x_i^+, x_i^-, h_i: i\in I\}$ such that $h_i = [x_i^+,x_i^-]$ and the subalgebra generated by $x_i^\pm, h_i$ is isomorphic to $\mathfrak{sl}_2$. Are the following two definitions equivalent?

A $\mathfrak g$-module $V$ is said to be locally finite-dimensional if every $v\in V$ lies in a finite-dimensional $\mathfrak g$-submodule.


A $\mathfrak g$-module $V$ is said to be integrable if it is a weight module (ie, $V = \bigoplus_{\lambda \in \mathfrak h^*} V_\lambda)$ and if the Chevalley generators $x_i^\pm$   act locally nilpotently on $V$ (i.e, $(x_i^\pm)^kv = 0$ for some $k\in \mathbb Z_{>0}$ and every $v\in V)$.

I'll show my work.
Integrable are locally finite-dimensional: Given $V$ integrable and $v\in V$, regarding the expression for the generators of $U(\mathfrak g$) via the PBW theorem, since $x_i^\pm$ are locally nilpotent and $V$ is a weight module, there are only a finite number of basis elements $x\in U(\mathfrak g)$ such that $xv \neq 0$. Thus, $U(\mathfrak g)v$ is finite-dimensional.
Locally finite-dimensional are integrable: Let $\{v_i\}$ be a basis of $V$. Since $v_i$ lies in a finite-dimensional $\mathfrak g$-submodule $V_i\subseteq V$ which is completly reducible, we see that $V$ is isomorphic to a sum of simple finite-dimensional $\mathfrak g$-submodules. Each of these simple finite-dimensional submodules are weight modules, so $V$ is a sum of weight spaces (will it be a direct sum?). Also, $x_i^\pm$ must act locally nilpotently, because the submodules are all finite-dimensional.
Is this proof fine? Am I missing something?
 A: 
... there are only a finite number of basis elements $x∈U(\mathfrak{g})$
such that $xv \neq 0$.

This is not true. If $ 0 \neq v \in V_\lambda$ and $\lambda$ is nonzero, there exists $h_i$ such that $\lambda(h_i) \neq 0$. Then $h_i^k.v = \lambda(h_i)^k v\neq 0$ for all $k\geq 0$.
But your idea of the proof that integrable modules are locally finite-dimensional still works. Let $v \in V$ be given. Write the subalgebra generated by $\{x_i^{\pm} \}$ as $\mathfrak{n}^{\pm}$. By the PBW theorem, $U(\mathfrak{g})v=U(\mathfrak{n}^+)U(\mathfrak{n}^-)U(\mathfrak{h})v$. If $v = v_1 + \cdots + v_l$ for $v_j \in V_{\lambda_j}$, then $U(\mathfrak{h})v \subset \text{span} \{v_1, \cdots, v_l \}$ so $U(\mathfrak{h})v$ is finite-dimensional. Since $x_i^\pm$ are locally nilpotent, $U(\mathfrak{n}^+)U(\mathfrak{n}^-)U(\mathfrak{h})v$ is also finite-dimensional. The followings are sketch of the details:
Put $W = U(\mathfrak{h})v$. Then $U(\mathfrak{n}^-)U(\mathfrak{h})v=U(\mathfrak{n}^-)W$ is the linear span of $\{y_1^{i_1} \cdots y_n^{i_n} w_k  : i_1, \cdots, i_n \geq 0, 1 \leq k \leq m\}$, where $\{w_1, \cdots, w_m \}$ is a basis for $W$ and $y_j=x_j^{-}$. Since $y_j$'s are locally nilpotent,  there exists a positive integer $M_n$ such that $i_n \geq M_n$ implies $y_n^{i_n}w_k = 0 $ for all $1 \leq k \leq m$. Thus the following set is finite. $$\{ y_n^{i_n} w_k  : i_n \geq 0, 1 \leq k \leq m\}$$
Repeating this process, we can see $\{y_1^{i_1} \cdots y_n^{i_n} w_k  : i_1, \cdots, i_n \geq 0, 1 \leq k \leq m\}$ is finite. Thus $\dim \Big( U(\mathfrak{n}^-)U(\mathfrak{h})v \Big) < \infty $. Similarly, we can conclude further that $$\dim U(\mathfrak{g})v = \dim \Big(U(\mathfrak{n}^+)U(\mathfrak{n}^-)U(\mathfrak{h})v \Big)<\infty$$

The converse part seems good. Note that if $V$ is a sum of weight spaces, then the sum would be direct. Suppose $V = \sum_{\lambda \in \mathfrak{h}^*}V_\lambda$. If the sum is not direct, there exists $v_1, \cdots , v_m \in V$ with $v_1 + \cdots + v_m = 0$ such that $0 \neq v_i \in V_{\lambda_j}$ and all $\lambda_j$ are distinct. We may assume $m \geq 1$ is minimal (In fact, $m$ cannot be $1$ so $m \geq 2$). Then there exists $h \in \mathfrak{h}$ such that $\lambda_1(h) \neq \lambda_2(h)$.
Now $$h.(v_1 + \cdots + v_m) = \lambda_1(h)v_1 + \cdots + \lambda_m(h)v_m=0$$ On the other hand, $$\lambda_1(h)(v_1 + \cdots + v_m) = \lambda_1(h)v_1 + \cdots + \lambda_1(h)v_m=0$$ Subtracting these two equalities, we get $$\left(\lambda_2(h) - \lambda_1(h) \right)v_2 + \cdots + \left(\lambda_m(h) - \lambda_1(h) \right)v_m $$ with $\lambda_2(h) - \lambda_1(h)  \neq 0$, which contradicts to the minimality of $m$.
