To expand on David's comment:  let $M_\lambda$ be the *lowest* weight Verma for $\lambda$ an integral weight, and consider $x\in M^*_{\lambda}$, we have a function on the Borel given by $f_x(b)=\langle \ell, b^{-1} x\rangle=\langle b\ell, x\rangle$  for $\ell$ the lowest weight vector in $M_\lambda$.  This gives a map $\oplus M^*_\lambda \to \mathbb{C}[B]$.  If we factor $b=nt$, then we see $f_x(b)=\lambda(t)f_x(n)$. The function $f_x(n)$ is independent of $\lambda$ (identifying all the lowest weight Verma modules as $\mathfrak{n}$-modules), so we can think of $x$ as lying in $U(\mathfrak{n})^*\cong \mathbb C[N]$ (here I'm taking restricted dual).   Since $\mathbb C[B]\cong \mathbb C[T]\otimes \mathbb C[N]$, we see that this gives that the map is injective and surjective.  This isomorphism is precisely set up so that the functions on $G/U_-$ are those which come from $x$ in a finite dimensional subrepresentation of $M_\lambda^*$.  

**EDIT:** And if you're wondering what that f.d. sub is, $M_\lambda$ has by assumption a lowest weight vector of weight $\lambda$, and so $M^*_{\lambda}$ has a highest weight vector of weight $-\lambda$.  Thus, it contains $V_{-\lambda}$ if $\lambda$ is anti-dominant, and no f.d. sub otherwise.