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In the paper [Jessen, B., Marcinkiewicz, J., and Zygmund, A. Note on the differentiability of multiple integrals. Fundamenta Mathematicae 25.1 (1935): 217-234] it is considered the limit $$ \lim_{\delta(I)\to0}|I|^{-1}\int_If(P)\,dP,\qquad\qquad\qquad (1) $$ where an integrable function $f$ is defined on a cube $S$, $I$ denotes any cell with sides parallel to the axes, contained in $S$, and containing some fixed point $P_0$, $|I|$ denotes the measure, and $\delta(I)$ the diameter of $I$.

In particular it is stated

Theorem 6. Let $\alpha_1(t), \alpha_2(t),\ldots, \alpha_n(t)$, be arbitrary non-decreasing functions defined to the right of $t=0$, vanishing and continuous for $t=0$, and positive for $t>0$. If the cells $I$ containing the point $P_0$ are of the form $$ \xi'_i\le x_i\le\xi''_i,\quad \xi'_i-\xi''_i=\alpha_i(t),\qquad i=1,2,\ldots,k, $$ then the limit (1) exists and is equal to $f(P_0)$ at almost every point $P(0)$.

So far as we are aware, this theorem has never been stated explicitly, although its proof is similar to that of the Lebesgue theorem mentioned at the beginning of this paragraph.

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We shall not give the proof of Theorem 6 here, for this would be a mere repetition of the usual proof of the Lebesgue theorem. It is sufficient to observe that the Vitali covering lemma, which plays the most fundamental part in the argument, remains valid for cells $I$ of the form just considered, and the proof is similar. We leave it to the reader to fill in the details of the proof.

Is this proof written down somewhere? I'm also interested in a stronger version, namely, that under the conditions of theorem 6 $$ \lim_{\delta(I)\to0}|I|^{-1}\int_I|f(P)-f(P_0)|\,dP=0 $$ for almost all points $P_0\in S$.

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The form of those collections $\{I\}$ for which the LDTh holds, are studied in several textbooks on Geometric Measure Theory. However, the usual assumption is bounded eccentricity (which does not seem to be sufficient here), and I don't have reference handy for Theorem 6. Yet "bounded eccentricity" + "Lebesgue differentiation theorem" may be valid keywords for a google search on other extension of the LDTh.

Thinking at the usual proof via the Vitali covering lemma, as e.g. in Wheeden and Zygmund's Measure and Integral: An Introduction to Real Analysis it seems to me everything works with no or little changes.

As to the stronger version of the limit, it follows plainly from theorem 6, again exactly as in the case of cubes from the usual Lebesgue differentiation theorem (in which case reads: "almost every point is a Lebesgue point"). You can find the easy proof in the above-mentioned book.

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  • $\begingroup$ Thank you for your answer. It's nice to know that the stronger version follows immediately. And yes, in all the sources I managed to find bounded eccentricity was supposed for $p=1$. $\endgroup$ – Andrew Jan 22 '15 at 7:48
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It may be derived from Lebesgue differentiation in dimension 1 by indicting in dimension. See, for example

Differentiation of Integrals in $\mathbb{R}^n$ (Lecture Notes in Mathematics) by M. de Guzman.

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