[proofplan] We first show that a submodule $N\subset M(\lambda)$ is closed under taking weight components. This is done by using polynomial projectors in $U(\mathfrak h)$ on the finitely many weight components of a chosen vector. Once $N$ contains weight vectors, we choose one whose weight is as high as possible among the weights occurring in $N$, measured by minimal height below $\lambda$. Positive-root operators raise weights, so applying any element of $\mathfrak n^+$ to this vector would produce a strictly higher weight in $N$ unless the result is zero. [/proofplan]

[step:Decompose vectors in $N$ into weight components that still lie in $N$]Let $Q_+$ denote the positive root monoid generated by the simple roots: \begin{align*} Q_+ := \sum_{\gamma\in\Delta}\mathbb Z_{\ge 0}\gamma. \end{align*} Define the height map \begin{align*} \operatorname{ht}:Q_+&\to\mathbb Z_{\ge 0} \end{align*} by declaring, for the unique expression $\beta=\sum_{\gamma\in\Delta}m_\gamma\gamma$ with $m_\gamma\in\mathbb Z_{\ge 0}$, \begin{align*} \operatorname{ht}(\beta):=\sum_{\gamma\in\Delta}m_\gamma. \end{align*} For a [Lie algebra](/page/Lie%20Algebra) $\mathfrak a$, write $U(\mathfrak a)$ for its universal enveloping algebra. By the PBW model of Verma modules [citetheorem:9374], every vector of $M(\lambda)$ is a finite sum of weight vectors, and every weight of $M(\lambda)$ has the form $\lambda-\beta$ for some $\beta\in Q_+$. We prove that if $v\in N$ has weight decomposition \begin{align*} v=v_1+\cdots+v_m \end{align*} where each $v_j$ is a nonzero weight vector of weight $\mu_j\in\mathfrak h^*$ and the weights $\mu_1,\dots,\mu_m$ are distinct, then each $v_j$ lies in $N$. Fix an index $j\in\{1,\dots,m\}$. For each $i\ne j$, since $\mu_i\ne\mu_j$ as elements of $\mathfrak h^*$, choose an element $h_i\in\mathfrak h$ such that \begin{align*} \mu_j(h_i)\ne\mu_i(h_i). \end{align*} Define an element $p_j\in U(\mathfrak h)$ by \begin{align*} p_j:=\prod_{i\ne j}\frac{h_i-\mu_i(h_i)}{\mu_j(h_i)-\mu_i(h_i)}. \end{align*} The denominators are nonzero by construction. Since $v_i$ has weight $\mu_i$, the element $h_i-\mu_i(h_i)$ kills $v_i$. Since $v_j$ has weight $\mu_j$, the factor indexed by $i$ acts on $v_j$ as multiplication by $1$. Therefore \begin{align*} p_jv=v_j. \end{align*} Because $N$ is a $\mathfrak g$-submodule, it is stable under the action of $U(\mathfrak g)$, hence under $U(\mathfrak h)\subset U(\mathfrak g)$. Since $v\in N$, this gives $v_j=p_jv\in N$.[/step]

[guided]The point of this step is to justify that we may look for a special vector in $N$ among weight vectors, not just among arbitrary vectors. Let \begin{align*} Q_+ := \sum_{\gamma\in\Delta}\mathbb Z_{\ge 0}\gamma \end{align*} be the positive root monoid generated by the simple roots. Define the height map \begin{align*} \operatorname{ht}:Q_+&\to\mathbb Z_{\ge 0} \end{align*} as follows: if $\beta=\sum_{\gamma\in\Delta}m_\gamma\gamma$ is the unique expression of $\beta$ in the simple-root basis with $m_\gamma\in\mathbb Z_{\ge 0}$, then \begin{align*} \operatorname{ht}(\beta):=\sum_{\gamma\in\Delta}m_\gamma. \end{align*} For a Lie algebra $\mathfrak a$, write $U(\mathfrak a)$ for its universal enveloping algebra. By the PBW description of Verma modules [citetheorem:9374], every vector of $M(\lambda)$ is a finite sum of weight vectors, and the possible weights are of the form $\lambda-\beta$ with $\beta\in Q_+$. Now take an element $v\in N$ and write its finite weight decomposition as \begin{align*} v=v_1+\cdots+v_m, \end{align*} where $v_j$ is a nonzero weight vector of weight $\mu_j\in\mathfrak h^*$, and the weights $\mu_1,\dots,\mu_m$ are pairwise distinct. We need to prove that each component $v_j$ still lies in $N$. The problem is that $N$ is known to be stable under $\mathfrak g$, not automatically under taking summands in an abstract [direct sum](/page/Direct%20Sum) decomposition. We isolate each summand using an operator built from $\mathfrak h$. Fix $j\in\{1,\dots,m\}$. For every $i\ne j$, the weights $\mu_i$ and $\mu_j$ are distinct linear functionals on $\mathfrak h$, so there exists $h_i\in\mathfrak h$ such that \begin{align*} \mu_j(h_i)\ne\mu_i(h_i). \end{align*} Define \begin{align*} p_j:=\prod_{i\ne j}\frac{h_i-\mu_i(h_i)}{\mu_j(h_i)-\mu_i(h_i)} \end{align*} as an element of $U(\mathfrak h)$. The denominator in each factor is nonzero by the choice of $h_i$. This operator acts as a projector onto the $j$-th weight component. Indeed, if $i\ne j$, then the factor $h_i-\mu_i(h_i)$ kills $v_i$, because $h_i$ acts on the weight vector $v_i$ by the scalar $\mu_i(h_i)$. On the other hand, for $v_j$, the same factor acts by the scalar \begin{align*} \frac{\mu_j(h_i)-\mu_i(h_i)}{\mu_j(h_i)-\mu_i(h_i)}=1. \end{align*} Thus every factor in $p_j$ acts as the identity on $v_j$, while at least one factor kills each other component. Hence \begin{align*} p_jv=v_j. \end{align*} Finally, because $N$ is a $\mathfrak g$-submodule, it is stable under the action of $U(\mathfrak g)$. Since $U(\mathfrak h)\subset U(\mathfrak g)$, the element $p_j$ sends $N$ to itself. Therefore $v_j=p_jv\in N$.[/guided]

[step:Choose a nonzero weight vector in $N$ of minimal height below $\lambda$] Since $N\ne 0$, choose $0\ne v\in N$. By the previous step, at least one nonzero weight component of $v$ lies in $N$. Hence the set \begin{align*} S:=\{\beta\in Q_+ : N_{\lambda-\beta}\ne 0\} \end{align*} is nonempty, where \begin{align*} N_{\lambda-\beta}:=N\cap M(\lambda)_{\lambda-\beta}. \end{align*} The set \begin{align*} \{\operatorname{ht}(\beta):\beta\in S\}\subseteq\mathbb Z_{\ge 0} \end{align*} is a nonempty subset of the nonnegative integers, so it has a least element. Choose $\beta_0\in S$ with minimal height, and choose \begin{align*} 0\ne w\in N_{\lambda-\beta_0}. \end{align*} Then $w$ is a nonzero weight vector of weight $\lambda-\beta_0$. [/step]

[step:Show every positive root vector kills the minimal-height vector] For $\alpha\in\Phi$, let \begin{align*} \mathfrak g_\alpha:=\{y\in\mathfrak g:[h,y]=\alpha(h)y\text{ for every }h\in\mathfrak h\} \end{align*} denote the corresponding root space. Let $\alpha\in\Phi^+$ and let $x\in\mathfrak g_\alpha$. Since $N$ is a $\mathfrak g$-submodule and $w\in N$, we have $xw\in N$. By the root-operator weight-shift property [citetheorem:9362], the vector $xw$ either is zero or has weight \begin{align*} (\lambda-\beta_0)+\alpha=\lambda-(\beta_0-\alpha). \end{align*} If $xw\ne 0$, then $\beta_0-\alpha\in Q_+$ because every weight of $M(\lambda)$ is of the form $\lambda-\beta$ with $\beta\in Q_+$. Because $\alpha\in\Phi^+$ is a positive integral combination of simple roots, $\operatorname{ht}(\alpha)>0$. Therefore \begin{align*} \operatorname{ht}(\beta_0-\alpha)=\operatorname{ht}(\beta_0)-\operatorname{ht}(\alpha)<\operatorname{ht}(\beta_0), \end{align*} contradicting the minimality of $\operatorname{ht}(\beta_0)$ among weights occurring in $N$. Therefore $xw=0$ for every $\alpha\in\Phi^+$ and every $x\in\mathfrak g_\alpha$. [/step]

[step:Conclude that the chosen vector is singular] The positive nilpotent subalgebra is \begin{align*} \mathfrak n^+=\bigoplus_{\alpha\in\Phi^+}\mathfrak g_\alpha. \end{align*} The previous step shows that every element of every summand $\mathfrak g_\alpha$ kills $w$. By linearity of the action, every element of $\mathfrak n^+$ kills $w$, so \begin{align*} \mathfrak n^+w=0. \end{align*} Since $w\ne 0$ and $w$ is a weight vector, $w$ is a nonzero singular vector contained in $N$. This proves the theorem. [/step]