integers $(x_i)_{i=1}^{k-1}$ with $x_i \in S_i$, such that $- \sum_i x_i \in S_k$.

Up to an error which is at most $O_k((s_1 + \dots + s_k)^{k-2})$, this can be written as an integral $$\int_{-s_1/2}^{s_1/2} \dots \int_{-s_{k-1}/2}^{s_{k-1}/2} {\mathds{1}}_{|\lfloor y_1\rfloor + \dots + \lfloor y_{k-1}\rfloor | {\leqslant}s_k/2}dy_1 \dots dy_{k-1}.$$ The two implications above allow us to show that, up to acceptable error, this is equal to $$\int_{-s_1/2}^{s_1/2} \dots \int_{-s_{k-1}/2}^{s_{k-1}/2} {\mathds{1}}_{|y_1+ \dots + y_{k-1} | {\leqslant}s_k/2}dy_1 \dots dy_{k-1},$$ which is equal to $\Phi(s_1, \dots, s_k)$; we omit the details.

We are now ready to prove Proposition \[prop:twelfth\_k\_var\].

As in Proposition \[prop:twelfth\], we will consider $S$ as the union of $k$ sets $S_1, \dots, S_k$, with the property that $T(S_1, \dots, S_k)$ is large.

The way we will do this is as follows. Let $M$ be a large integer, which we assume to be divisible by $2k$. Define $$S_i = \frac{1}{a_i}[-M/2k, M/2k]$$ for each $i$ with $1 {\leqslant}i {\leqslant}k$, where we may normalise the sets to consist of integers by multiplying by $\prod_i a_i$. Then, let $S = \cup_i S_i$, so that $|S| {\leqslant}M$.

It remains to show that $T(S_1, \dots, S_k) {\geqslant}\frac{\sigma_k}{k^{k-1}} M^{k-1} + O(M^{k-2}).$ But this follows as an easy consequence of 