Let $K$ be a finite extension of $\mathbb{Q}\_p$ and $G$ be a split connected reductive algebraic group over $K$ with Borel $B$. We have the associated Lie algebras $\mathfrak{g}=$Lie$(G)$ and $\mathfrak{b}=$Lie$(B)$. 


Let $M$ be a $U(\mathfrak{g})$-module with $N \subset M$ a finite dimensional $K$-module, which is $B$-invariant and generates $M$ as a $U(\mathfrak{g})$-module. 


I read that $M$ is then locally $\mathfrak{b}$-finite, i.e $U(\mathfrak{b}) \cdot m \subset M$ is finite dimensional for all $m \in M$, but I have trouble to see this. As $U(\mathfrak{b})$ seems so big for me, I cannot think of a finite basis for $U(\mathfrak{b}) \cdot m \subset M$ by knowing only $N$.



Checking axiom of Category $\mathcal{O}$
The module $U(\mathfrak{g})\otimes\_{U(\mathfrak{b})} N$ is locally $U(\mathfrak{b})$-finite and there is a surjective $U(\mathfrak{g})$-homomorphism $\varphi\colon U(\mathfrak{g})\otimes\_{U(\mathfrak{b})} N \to M$ given by $u \otimes n \mapsto u\cdot n.$ It is easy to see that every weight space of $M$ has only finitely-many preimages. It follows that $M$ cannot contain infinite-dimensional $U(\mathfrak{b})$-module. In the case $N$ is one-dimensional, the module I just constructed is called Verma module. In case of $N$ not being completely reducible, one would have to do some gymnastics with short exact sequences. 


But perhaps for a beginner it's easier to just attack the problem directly. Assume first for simplicity that $m = u \cdot n$ for some $u \in \mathfrak{g}.$ Pick $X \in \mathfrak{b}.$ Then
$$
X \cdot m = X\cdot u \cdot n = Xu \cdot n = [X,u]\cdot n + uX\cdot n = u'\cdot n + u \cdot n',
$$
 where $u'$ is some other element of $\mathfrak{g}$ and $n'$ is some other element of $N.$ Now iterate for general $u\in U(\mathfrak{g})$ and repeat for all possible $X \in \mathfrak{b}$. You will see that you pick up at most $\dim \bigotimes^k \mathfrak{g} \otimes N$ possible elements. They might not be all linearly independent, but they surely span $\mathfrak{b} \cdot m.$


