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Lemma. Suppose $g$ is a continuous function on an open set $U$ in $\mathbb{R^2}$, $p=(x_0,y_0) \in U$ and let $Q_{n}=[x_0-\epsilon_n,x_0+\epsilon_n] \times [y_0-\epsilon_n,y_0+\epsilon_n]$ be a sequence of rectangles converging to $p$. Then $$\lim_{n \rightarrow \infty} \frac{1}{\text{vol}(Q_n)}\int_{Q_n}g=g(p).$$

I was looking at a proof and this lemma was pulled out. There is this question where someone asked about what seems to be a more general version of the same fact but it uses some stuff beyond the scope of multivariable calculus so I was wondering if there is a simpler way to prove this specific version of it.

The "volume" for $Q_{n}$ in this notation would be $(2\epsilon)^{2}$, so we may write this limit as $$\lim_{n \rightarrow \infty} \frac{1}{(2\epsilon_{n})^2}\int_{Q_n}g$$ and $\epsilon_{n}$ approaches $0$ since the rectangle is converging to $p$, but I can't think of a way to manipulate the integral so I can prove the equality. Any hints and nudges in the right direction would be very much appreciated.

gr8astard
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1 Answers1

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Hint:

Since $g$ is a continuous function, for any $\delta \gt 0$, there exists an $\epsilon \gt 0$ such that

$$|g(x,y) - g(p)| \lt \delta \; \forall \; (x,y) \in Q = [x_0-\epsilon,x_0+\epsilon] \times [y_0-\epsilon,y_0+\epsilon] \tag{1}\label{eq1A}$$

Of course, for any $\epsilon$ which works, you can use any smaller $\epsilon \gt 0$. Note that \eqref{eq1A} gives

$$\int_{Q}|g(x,y) - g(p)| \lt (2\epsilon)^2\delta \tag{2}\label{eq2A}$$

For example, you don't mention the codomain of $g$ but, for simplicity, if it's $\mathbb{R}$, you then have from \eqref{eq1A} that

$$-\delta \lt g(x,y) - g(p) \lt \delta \tag{3}\label{eq3A}$$

in which case you can rewrite \eqref{eq2A} as

$$\begin{equation}\begin{aligned} -(2\epsilon)^2\delta & \lt \int_{Q}(g(x,y) - g(p)) \lt (2\epsilon)^2\delta \\ -(2\epsilon)^2\delta & \lt \int_{Q}(g(x,y)) - (2\epsilon)^2g(p) \lt (2\epsilon)^2\delta \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

I trust you can finish the rest yourself (e.g., by using something like the Squeeze theorem).

John Omielan
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  • I think I got it. Going forward from (2), I can bound the absolute difference between $g(p)$ and $\frac{1}{(2\epsilon)^2}\int_{Q_n}g$ by an arbitrarily small $\delta$ thanks to the continuity of $g$, since there will be some $N$ such that for $n \ge N$, our rectangle $Q_n$ will be small enough. – gr8astard Mar 20 '20 at 07:45
  • @gr8astard Yes, I believe you do have it. I left those final details to you because I thought it would be better for you to figure them out yourself as a learning experience. – John Omielan Mar 20 '20 at 07:47
  • Ok I actually got it and commented just as you edited the rest of the details. I did get there in the end haha thank you – gr8astard Mar 20 '20 at 07:52