Convergence of sum to integral - unit ball Let $S$ be the set $\{(x,y)\in \mathbb{R}^2:x^2+y^2\leq 1\}$ and $f(k):S \rightarrow \mathbb{R}$ a continuous function.
I would like to prove
$$
\lim_{T \rightarrow \infty}\frac{1}{T^2}\sum_{k \in \frac{\mathbb{Z}^2}{T} \cap S} f(k) = \int_S f(s) ds.
$$
The image below suggests that the statement is true. Can someone give me a hint how to do the formal proof.

Idea:
It is well known that for $L>0$ the Riemann sum
$$
\frac{1}{T^2}\sum_{k\in \frac{[-L,L]^2}{T}}f(k)
$$
converges for $T\rightarrow \infty$ against the integral
$$
\int_{[-L,L]^2}f(x)dx
$$
for all continuous functions on $[-L,L]^2$. For sufficiently large $L$ we consider the restricted space $C([-L,L]^2\lvert_S)$ of continuous functions. [Throw everything away that is outside the ball S.] I guess $C([-L,L]^2\lvert_S)=C(S)$ and therefore
$$
\lim_{T \rightarrow \infty}\frac{1}{T^2}\sum_{k \in \frac{\mathbb{Z}^2}{T} \cap S} f(k) = \int_S f(s) dx
$$
for all continuous functions.
 A: This is a standard thing with multivariate integrals, as long as the boundary is nice enough, you may throw away the boundary ''area'' in your integral. In particular, you may integrate over non-rectangular ``decent'' shapes.
Notice that the boundary of the disk, namely the unit sphere - has what is called area zero. Moreover, we can make it slightly more quantitative.
Consider $Bad_{T}$ to be the squares of side-length $1/T$, positioned over the lattice $\mathbb{Z}^2/T$ which intersect the boundary of $S$.
Denote $Good_{T}$ to be the set of such squares which are contained inside $S$.
It is clear that $Good_{T}\subset S \subset Good_{T}\cup Bad_{T}$.
[you can be a bit more pedantic with the definition of bad, say focusing on the bottom left corner if you wish to get your sketch on the nose, it won't even matter, it is just going to effect the error term by a constant].
I claim that $\sum area(Bad_{T})$ is $O(T^{-1})$. Hence by naïve sup bound over $f$, we can safely ignore this part of the summation, and just focus on the summation over the ``Good'' part, which is clearly dominated by the integral over $S$ (yes you want to use uniform continuity and once you fix $\varepsilon>0$, take $T$ such that $T^{-1}<\varepsilon/\sqrt{2}$ or so, just to be on the safe side and blur the scale enough).
On the other hand, it is clear you over estimate the integral over $S$, hence a squeeze argument finishes things up.
In order to prove the assertion about the bad squares, we are essentially interested in the following count - how many lattice points of $\mathbb{Z}^2/T$ are in the shell between $1-\sqrt{2}/T$ and $1+\sqrt{2}/T$, or equivalently, to a less confusing scaling, lattice points of $\mathbb{Z}^2$ with norm between $T-\sqrt{2}$ to $T+\sqrt{2}$.
Evidently, as the lattice points is discrete and any unit square contains $O(1)$ lattice points, just from area considerations we see that we have $O((T+\sqrt{2})^2-(T-\sqrt{2})^2) = O(4\sqrt{2}T)$ such squares, where I didn't take any pain in optimization here. Now going back to the scaled version, we need to multiply each square by its area ($T^{-2}$), leading to the desired estimate of $O(T^{-1})$.
