Finite-grid masked sampler defines a normalized law

For $j\ge 1$, strict decrease gives ${\alpha }_j<1$ and ${\alpha }_{j-1}>{\alpha }_j\ge 0$, so $c_j>0$. Since ${\alpha }_{j-1}\le 1$, ${\alpha }_{j-1}-{\alpha }_j\le 1-{\alpha }_j$, hence $c_j\le 1$.

If $z_i\mathrm{\ne }m$, the coordinate row in [F1] is a point mass and sums to one. If $z_i=m$, its mass is $(1-c_j)+{\sum }_{v\in V}{c}_j{h}_{i,j}(z)(v)=1$ because $h_{i,j}(z)$ is a probability vector. The product over positions therefore has nonnegative entries summing to the product of $L$ ones, namely one; by [F2] it is a probability-kernel row.

Fix a current state $z$ and position $i$. If $z_i=m$ and $c_j<1$, the uniform threshold satisfies $\mathrm{Pr}(U_{j,i}^{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{k}}\ge c_j)=1-c_j$ and $\mathrm{Pr}(U_{j,i}^{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{k}}<c_j)=c_j$. For an ordered vocabulary $v_1,\dots ,v_K$, put $H_0=0$ and $H_k={\sum }_{a=1}^k{h}_{i,j}(z)(v_a)$. The inverse-CDF rule selects $v_k$ on the interval $[H_{k-1},H_k)$, whose uniform probability is $H_k-H_{k-1}=h_{i,j}(z)(v_k)$. The fallback at $U_{j,i}^{\mathrm{t}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{n}}=1$ has probability zero and so does not change this law, even if the fallback token has zero mass. Since the mask and token uniforms are independent, the resulting row assigns mass $1-c_j$ to $m$ and $c_j{h}_{i,j}(z)(v)$ to each $v$. If $c_j=1$, the direct token draw has law $h_{i,j}(z)$ and no mask is retained. Unmasked coordinates are copied, and independence of the random pairs across positions makes the sequence row the product in [F1]. Thus the stated random instructions realize the coordinate and product rows from step 2.1.

The loop has exactly $N$ reverse-step iterations, each updates the finite set of $L$ positions once, and each inverse-CDF search checks at most the finite number $K=\mathrm{\mid }V\mathrm{\mid }$ of tokens; the fallback defines the draw also at $U_{j,i}^{\mathrm{t}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{n}}=1$. Hence every instruction terminates and the algorithm terminates after finitely many operations. Starting from the point mass at $m^L$, composition of the normalized rows from step 2.1 gives a normalized finite path and endpoint law. At $j=1$, $c_1=({\alpha }_0-{\alpha }_1)\mathrm{/}(1-{\alpha }_1)=1$, so the algorithm's direct token-draw branch removes every remaining mask and copies every already clean coordinate. Thus $Z_0\in V^L$, completing the support claim.