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I wondered whether it holds, for a function $f:A\subset\mathbb{R}^3\to\mathbb{R}$, $f\in C(A)$, that, for all $(x_0,y_0,z_0)\in A$, $$\lim_{\sqrt{h_x^2+h_y^2+h_z^2}\to 0}\frac{1}{h_x h_y h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz=f(x_0,y_0,z_0)$$I think that, if such an identity held, it would allow us to define quantities like the densities used in physics.

Obviously, for all $\varepsilon>0$ there exists a $\delta$ such that, if $\sqrt{h_x^2+h_y^2+h_z^2}<\delta$ (and therefore $|h_x|,|h_y|,|h_z|$ $<\delta$), then $$\Bigg|\frac{1}{h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz-\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z_0)dxdy\Bigg|<\varepsilon$$$$\Bigg|\frac{1}{h_y}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z_0)dxdy-\int_{x_0}^{x_0+h_x}f(x,y_0,z_0)dx\Bigg|<\varepsilon$$$$\Bigg|\frac{1}{h_x}\int_{x_0}^{x_0+h_x}f(x,y,z_0)dxdydz-f(x_0,y_0,z_0)\Bigg|<\varepsilon$$but I am not able to use these inequalities to find an arbitrarily small $\tilde{\varepsilon}$ to majorate $\big|\frac{1}{h_xh_yh_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz-f(x_0,y_0,z_0)\big|$ with.

Does what I am trying to prove hold and, if it does, how can it be proved? I $\infty$-ly thank anybody for any answer.

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    $\begingroup$ You might want to look up Lebesgue's differentiation theorem. $\endgroup$ – Mike Sep 8 '15 at 7:45
  • $\begingroup$ to make that integral well-defined you should assume that $(x_0, y_0, z_0)$ is an interior point and exclude $h_xh_yh_z = 0$. $\endgroup$ – user251257 Sep 8 '15 at 13:27
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You don't need Lebesgue density / differentiation theorem, as $f$ is continuous.

As I am lazy, I will prove it in 2D and assume without loss of generality $x_0 = y_0 = 0$ and $f(0,0) = 0$. We need to show that $$ \lim_{r\to 0} \sup_{\|h\| < r, h_xh_y\ne 0} \left | \frac1{h_x h_y} \int_0^{h_y} \int_0^{h_x} f(x,y) \; \mathrm dx\mathrm dy \right | = 0.$$

Now, let $\epsilon > 0$. Then, by continuity of $f$ there is some $r > 0$ such that for every $h$ with $\|h\|< r$ it follows $|f(h_x,h_y)| < \epsilon.$ Notice that for every $(x,y)$ with $|x| \le |h_x|$ and $|y|\le |h_y|$ we have $$ \sqrt{x^2 + y^2} \le \| h \| < r. $$ Thus, for $h_xh_y \ne 0$ and $s_x = \operatorname{sign}(h_x), s_y = \operatorname{sign}(h_y)$ we obtain \begin{align} \left | \frac1{h_x h_y} \int_0^{h_y} \int_0^{h_x} f(x,y) \; \mathrm dx\mathrm dy \right | &= \frac1{|h_x| |h_y|} \left | \int_0^{|h_y|} \int_0^{|h_x|} f( s_x x, s_y y) \; \mathrm dx\mathrm dy \right | \\ &\le \frac1{|h_x| |h_y|} \int_0^{|h_y|} \int_0^{|h_x|} \left | f( s_x x, s_y y)\right | \; \mathrm dx\mathrm dy \\ &< \frac1{|h_x| |h_y|} \int_0^{|h_y|} \int_0^{|h_x|} \epsilon \; \mathrm dx\mathrm dy \\ &= \epsilon. \end{align}

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  • $\begingroup$ Thank you so much! The Heine-Borel theorem allows us to see that, if $\|h\|$ is small enough for a closed ball $\bar{B}((x_0,y_0,z_0),\|h\|)$ to be contained in $A$, then $f-f(x_0,y_0,z_0)$ is uniformly continuous and therefore for all $\varepsilon>0$ there is a $\delta$ such that, if $\|h\|<\delta$, for all $(x,y,z)\in A$ such that $\|(x,y,z)-(x_0,y_0,z_0)\|\le \|h\|$, $|f(x,y,z)-f(x_0,y_0,z_0)|<\varepsilon$ and (I write the endpoint of the integrals as if $h_x,h_y, h_z>0$ but reversing then when they are negative doesn't change the absolute value) therefore... $\endgroup$ – Self-teaching worker Sep 8 '15 at 16:24
  • $\begingroup$ ...$\big|\frac{1}{h_x h_y h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz-f(x_0,y_0,z_0)\big|$ $=\big|\frac{1}{h_x h_y h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)-f(x_0,y_0,z_0)dxdydz\big|$ $\le \frac{1}{|h_x h_y h_z|}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}|f(x,y,z)-f(x_0,y_0,z_0)| dxdydz $ $\le \frac{1}{|h_x h_y h_z|}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}\varepsilon dxdydz$ $=\varepsilon$. $\endgroup$ – Self-teaching worker Sep 8 '15 at 16:24
  • $\begingroup$ Interestingly, I would say that the same applies to spaces of any dimension, so that we can say that, if $\boldsymbol{x}_0\in\mathring{A}$, then $$\lim_{\|\boldsymbol{h}\| \to 0,\prod_{i=1}^n h_i\ne 0}\frac{1}{\prod_{i=1}^n h_i}\int_{x_{0,n}}^{x_{0,n}+h_n}...\int_{x_{0,n}}^{x_{0,1}+h_1}f(\boldsymbol{x})dx_1...dx_n=f(\boldsymbol{x_0})$$ Thank you so much again!!! $\endgroup$ – Self-teaching worker Sep 8 '15 at 16:24
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    $\begingroup$ @Self-teachingDavide: You don't need Heine-Borel for that. It is just continuity at $(x_0,y_0,z_0)$. Yes, it applies to any finite dimension. $\endgroup$ – user251257 Sep 8 '15 at 16:26
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    $\begingroup$ @Self-teachingDavide: sure. $\endgroup$ – user251257 Sep 8 '15 at 16:53

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