Weights of BGG dual of Verma module Consider the Verma module $M(\lambda)=\oplus_{\mu \in \mathfrak{h}^{\star}} M(\lambda)_{\mu}$. Denote its BGG dual by $M(\lambda)^{\vee}=\oplus_{\mu \in \mathfrak{h}^{\star}} M(\lambda)^{\star}_{\mu}$ as in section $3.2$ of Humphreys' book.
When $\lambda$ is anti-dominant, $M(\lambda)$ is simple and $M(\lambda) \cong M(\lambda)^{\vee}$. My question is whether
$$\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{-\mu}?$$
I have doubts because on one hand, I feel that since $M(\lambda) \cong M(\lambda)^{\vee}$ we should have $\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{\mu}$. On the other hand, I feel that the weights of $M(\lambda)^{\vee}$ should be negative of the weights of $M(\lambda)$. Hence, it might be true that $\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{-\mu}$. 
Thank you for explaining.
 A: If you compute the dual of a weight module $M$ using the usual notion of dual of a module over a Lie algebra, then you get a module $M^{\ast}$ such that $M^{\ast}_{-\mu}$ is the vector space dual of $M_{\mu}$ (and thus has the same dimension).  This why you think the displayed equation is correct.
But that's not the notion of dual that anyone uses in category $\mathcal{O}$, because as you can see it doesn't send objects in category $\mathcal{O}$ to objects in category $\mathcal{O}$.  Thus, $M^\vee$ is something different: it's the usual vector space dual (EDIT: not actually the full one; as the OP writes, it's the sum of the vector space duals of the individual weight spaces, which is a version of graded dual), with the action twisted by the Cartan involution $E_i\mapsto -F_i, F_i\mapsto -E_i, H_i\mapsto -H_i$ (for the classical groups, this is transpose and negate).  Because of the negative sign in front of $H_i$, this flips the sign of weights, and $M^\vee_\mu$ is the vector space dual of $M_\mu$ (and thus the same dimension).  
