Hint to prove this lemma about "mean value of an integral" converging to function 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.
 A: 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).
