Q:

how to calculate limit of double integral

I have a rather simple problem which I can't make rigorous:
Let $\varphi \colon \mathbb{R}^2 \to \mathbb{R}$ be a continuous function. Then it holds that
$$ \lim_{m \to \infty} \, \, \,\lim_{n \to \infty} mn \int_{x}^{x+1/m} \int_y^{y+1/n} f(u,v) \, du dv = \lim_{m \to \infty} m \int_{x}^{x+1/m}  f(u,y) \, du =  f(x,y)$$.
Intuitively, it should also hold that
$$ \lim_{m \to \infty} m^2 \int_{x}^{x+1/m} \int_y^{y+1/m} f(u,v) \, du dv = f(x,y).$$
How can one prove this?

A:

Use the mean value theorems for definite integrals, and you should have
$$ \int_x^{x+1/m} \int_y^{y+1/m} f(u, v) \,dudv = \frac{1}{m^2} f(\alpha_m, \beta_m),$$ where $\alpha_m \in [x, x+1/m], \,\beta_m \in [y, y+1/m]$. 
The continuous property of $f(u, v)$ implies that
$$\lim_{m \to \infty} m^2 \int_x^{x+1/m} \int_y^{y+1/m} f(u, v) \,dudv = \lim_{m \to \infty} f(\alpha_m, \beta_m) = f(x, y).$$
References: The Mean Value Theorem for Double Integrals

