Do scaled-down integer lattice points serve as unbiased sample points in the probability simplex?

I'm stuck trying to prove a statement that seems intuitively obvious but not really straightforward to establish rigorously.

The $(n-1)$-dimensional probability simplex $X$ is the set $$ X=\left\{\mathbf{x}\in[0,1]^n:\sum_{i=1}^nx_i=1\right\}. $$ Let $U$ be the set of the scaled down integer lattice points in the simplex: $$ U=X\cap\frac{\mathbb{Z}^n}s $$ where $s\in\mathbb{Z}^+$.

My claim is that $U$ serves as an unbiased set of sample points in $X$ as $s\to\infty$. That is, for a subset $P\subseteq X$, its centroid (unweighted average) given by $$ \tag{1} \bar{\mathbf{p}}=\frac{1}{\int_{P}dV}\int_{P}\mathbf{p}\,dV $$ where $\mathbf{p}\in P$ and $dV$ is the $(n-1)$-dimensional volume element, is asymptotically equal to $$ \tag{2} \bar{\mathbf{u}}=\frac1{\#U}\sum_{i=1}^{\#U}\mathbf{u}_i $$ where $\mathbf{u}_i\in U$, as $s\to\infty$; i.e. $$ \tag{3} \lim_{s\to\infty}\bar{\mathbf{u}}=\bar{\mathbf{p}}. $$ Here, $(1)$ can be viewed as the true mean and $(2)$ as the sample mean, and $(3)$ holds if the sample points provided by the set $U$ are unbiased in the limit.

For my specific case, the statement doesn't have to hold for all subsets in $X$ but only for this particular set: $$ \tag{4} {P}=\left\{\mathbf{p}\in[0,1]^n: p_1\le p_2\le\cdots\le p_c\ge\cdots\ge p_n\text{ and } \sum_{i=1}^n p_i=1\right\}\subset X $$ which, by the way, happens to be convex.

In short, I'm trying to prove $(3)$, either in the general case or when $P$ is given as $(4)$.

Any help would be really appreciated. Guiding me to any related keyword or book would also be great.

It can't possibly hold for all subsets. Consider the subset of points with irrational coordinates. This has the same measure as all of $X$ but has no scaled-integer points in it for any $s$.

If the indicator function $I_P$ is Riemann integrable then every sequence of Riemann sums converges to the integral. Set up a hyper-rectangle grid with one of your scaled-integer points in each cell. You do now need to weight each grid cell by its volume, but all the cells that are not adjacent to the edge of the simplex have the same volume, and the fraction of grid cells adjacent to the edge goes to zero as $s$ increases.

Your set $P$ is not only convex but is the intersection of finitely many halfspaces, so its indicator function is definitely Riemann integrable: it's the product of finitely many step functions. I would guess that this is true for any convex set, but I don't have a proof.

I will finally note that the bias due to edge effects, while it goes to zero, is still asymptotically larger than the standard error of a Monte Carlo integral with the same number of points when $n>2$