Local nilpotence of only the ei does not imply integrability

Let $v$ have weight $\lambda -\beta$ with $\beta ={\sum }_j{b}_j{\alpha }_j\in Q^{+}$. Applying $e_i^{N}v$ gives a vector of weight $\lambda -\beta +N{\alpha }_i$. For $N=b_i+1$, its difference below $\lambda$ has $i$-coordinate $-1$, so it is not in $Q^{+}$ by independence of the simple roots. F2 implies that weight space is zero, hence $e_i^{b_i+1}v=0$. Every vector has finitely many weight components; taking one plus the maximum of their $i$-coefficients gives an exponent killing all components. For the zero vector take exponent one. Thus every $e_i$ is locally nilpotent on the entire module.

F1 supplies the same module's nonzero highest vector on which every power of $f_i$ is nonzero. It therefore fails integrability despite the verified raising condition. In the rank-one example at $\lambda =0$, the vector $f^k{v}_0$ has weight $-k\alpha$, so step 1.1 gives $e^{k+1}{f}^k{v}_0=0$, while $f^N{v}_0\mathrm{\ne }0$ for every $N$. For $k=0$ the raising bound is one; arbitrary finite sums use the maximum bound, not a global uniform bound. The simple index set must be nonempty for the failed lowering requirement to exist. This calculation is choice-free.