This is a problem that I recently saw in some lecture notes is proving a little more challenging that what I anticipated:
Suppose $f$ is a continuous function on $\mathbb{R}^2$ and such that$$f(x+1,y)=f(x,y)=f(x,y+1)\quad \text{for all}\quad(x,y)\in\mathbb{R}^2$$Show that$$\int_{[0,1]^2}f=\lim_{a\rightarrow\infty}\frac{1}{a}\int^a_0f(t,\alpha t)\,dt\tag{2}\label{two}$$whenever $\alpha\in\mathbb{R}\setminus\mathbb{Q}$.
My initial though was to consider the class of all functions satisfying (1) for which $\eqref{two}$ holds and then argue a complex version of the Stone-Wierstrass theorem.
However, I am having trouble proving the result for say, functions such as $\phi_{n,m}(x,y)=e^{2\pi i(nx+my)}$, where $n$, $m$ are integers. Maybe someone in the forum has a better idea or can work this out for $\phi_{n, m}$.
Edit: Thanks to Greg Martin for his comment which led me to realize of an obvious fact that I missed: $1$ and $\alpha$ are $\mathcal{Q}$-linearly independent.
For $\phi_{0,0}$ the statement is trivial. If $n^2+m^2>0$, then $n+m\alpha\neq0$ and
$$\begin{align}\frac{1}{a}\int^a_0 e^{2\pi i(n + \alpha m)t}\,dt&=\frac{1}{2\pi i(n+\alpha m)a}\big(e^{2\pi i (n+m\alpha)a}-1\big)\xrightarrow{a\rightarrow\infty}0\\\int_{[0,1]^2}e^{2\pi i (nx+my)}\,dxdy &=\Big(\int^1_0e^{2\pi i n x}\,dx\Big)\Big(\int^1_0e^{2\pi i m y}\,dy\Big)=0\end{align}$$
Thus $\eqref{two}$ holds for any trigonometric polynomial in $\mathbb{S}^1\times\mathbb{S}^1$. Since trigonometric polynomials are dense in $\mathcal{C}(\mathbb{S}^1\times\mathbb{S}^1)$ with the uniform norm, the result follows