Direct summands of Verma-filtered objects are Verma-filtered

If $m=0$ then $X=0$, so $X_1=X_2=0$ and both summands are Verma-filtered with the empty flag.

Let $m\ge 1$. The finite set of labels of the fixed flag has a maximal element $\nu$; every weight of $X$ lies below one of those labels, so a weight strictly above $\nu$ would force a flag label strictly above it. Thus $\nu$ is a maximal element of the set of weights of $X$, with no greatest-label assumption. By [F1] the weight $\nu$ is the label of some factor of the fixed flag: it lies in ${\mu }_i-Q^{+}$ for a factor $\mathrm{\Delta }({\mu }_i)$, so $\nu \le {\mu }_i$, while ${\mu }_i$ is a weight of $X$ and $\nu$ is maximal, so comparability $\nu \le {\mu }_i$ forces ${\mu }_i=\nu$. Choose a nonzero vector $v$ of weight $\nu$; writing $v=v_1+v_2$ with $v_i\in X_i$, some $v_i\mathrm{\ne }0$ is again a maximal-weight vector, and after exchanging the names of the summands we may assume $v\in X_1$. Assume as induction hypothesis that every Verma-filtered direct sum with a flag of length $m-1$ has Verma-filtered summands.

By [F2] the induced homomorphism $\mathrm{\Delta }(\nu )\to X$ is injective with image in $X_1$ (its image is the submodule generated by $v$) and cokernel Verma-filtered with a flag of length $m-1$. Since $\mathrm{\Delta }(\nu )\subseteq X_1$, [F1] gives $X\mathrm{/}\mathrm{\Delta }(\nu )\cong (X_1\mathrm{/}\mathrm{\Delta }(\nu ))\oplus X_2$; the induction hypothesis applies to this Verma-filtered direct sum of flag length $m-1$, so $X_2$ and $X_1\mathrm{/}\mathrm{\Delta }(\nu )$ are Verma-filtered. From the exact sequence $0\to \mathrm{\Delta }(\nu )\to X_1\to X_1\mathrm{/}\mathrm{\Delta }(\nu )\to 0$, both ends Verma-filtered, [F3] makes $X_1$ Verma-filtered.

By induction on $m$, steps 1.1 and 2.1 show that whenever $X=X_1\oplus X_2$ is Verma-filtered, both summands are Verma-filtered.