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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?

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  • $\begingroup$ Note that in the limit, you're essentially convolving $f$ with the top-hat representation of the Dirac delta function. $\endgroup$ – John Barber May 1 '18 at 2:21
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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

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  • $\begingroup$ Thanks, an argument like this was what I was looking for! $\endgroup$ – herrsimon May 1 '18 at 2:19
  • $\begingroup$ Why don't replace one $m$ with $n$? $\endgroup$ – Danny Pak-Keung Chan May 1 '18 at 2:32
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No. However intuitive you find it, it is simply NOT true that you can combine limits like that. Take the following as a counterexample:

$$\lim_{m \rightarrow \infty}\lim_{n \rightarrow \infty}\biggl(1+\frac 1n\biggl)^m=\lim_{m \rightarrow \infty}1^m=1$$

However,

$$\lim_{n \rightarrow \infty}\biggl(1+\frac 1n\biggl)^n=e$$

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  • $\begingroup$ What you say is true in general but your "counterexample" doesn't apply to the situation here. $\endgroup$ – herrsimon May 1 '18 at 2:18

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