Let $U \subseteq \mathbb{R}^{2} $ be some open set and $f \colon U \to \mathbb{R}$ a continuous function. Show that for any $\mathbf{p} \in U$ and a rectangle $R_{h} = \mathbf{p} + [-h, h]^{2}\subseteq U$ we have \begin{align*} \lim_{h\to 0} \frac{1}{4h^{2}}\int_{R_{h}}^{}\left| f( \mathbf{x}) - f( \mathbf{p})\right | \mathrm{~d}\mathbf{x} = 0 .\end{align*}
We have \begin{align*} \lim_{h\to 0} \frac{1}{4h^{2}}\int_{R_{h}}^{}\left| f( \mathbf{x}) - f( \mathbf{p})\right | \mathrm{~d}\mathbf{x} &= \lim_{h\to 0} \frac{1}{4h^{2}}\int_{[-h, h]^{2}}^{} \left| f( \mathbf{\mathbf{p} + \mathbf{h}}) - f( \mathbf{p})\right | \mathrm{~d}\mathbf{h} \\[5pt] &=\lim_{h\to 0} \frac{1}{4h^{2}}\int_{-h}^{h }\int_{-h}^{h} \left| f( \mathbf{\mathbf{p} + \mathbf{h}}) - f( \mathbf{p})\right | \mathrm{~d}h _{1} \mathrm{~d}h _{2} \end{align*} with $\mathbf{h} =\begin{bmatrix} h_{1} & h_{2} \end{bmatrix}^{\mathsf{T}} $.
I know that I somehow have to exploit continuity at this point but don't really see how I can proceed.
We have \begin{align*} \lim_{h\to 0} \frac{1}{4h^{2}}\int_{R_{h}}^{}\left| f( \mathbf{x}) - f( \mathbf{p})\right | \mathrm{~d}\mathbf{x} &= \lim_{h\to 0} \frac{1}{4h^{2}}\int_{[-h, h]^{2}}^{} \left| f( \mathbf{\mathbf{p} + \mathbf{h}}) - f( \mathbf{p})\right | \mathrm{~d}\mathbf{h} \\[5pt] &=\lim_{h\to 0} \frac{1}{4h^{2}}\int_{-h}^{h }\int_{-h}^{h} \left| f( \mathbf{\mathbf{p} + \mathbf{h}}) - f( \mathbf{p})\right | \mathrm{~d}h _{1} \mathrm{~d}h _{2} \end{align*} with $\mathbf{h} =\begin{bmatrix} h_{1} & h_{2} \end{bmatrix}^{\mathsf{T}} $.
Using the mean value theorem for integrals we can find some $a \in [-h, h]$ s.t. \begin{align*} \int_{-h}^{h} |f( \mathbf{p} + \mathbf{h})- f( \mathbf{p})| \mathrm{~d}h _{1} =2h |f(p _{1} + a, p _{2} + h_{2}) - f( \mathbf{p})| .\end{align*} Similarily we can find some $b \in [-h, h]$ s.t. \begin{align*} 2h\int_{-h}^{h} \left| f( p _{1} + a, p _{2} + h _{2}) - f( \mathbf{p}) \right | \mathrm{~d}h _{2} = 4h^{2} \left| f( p _{1} + a, p _{2} + b) - f( \mathbf{p}) \right | .\end{align*} The $4h^{2}$ in the numerator and denominator of the limit cancel out and using continuity of $f$ once again we find \begin{align*} \lim_{h\to 0} \left| f( p _{1} + a, p _{2} + b) - f( \mathbf{p}) \right | = 0 .\end{align*}