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This error is also commented in the Addenda of the book, with a proposed solution, but I don't understand how it should fix the proof.

The (part of the proof of the) theorem involved is:

Suppose $f:A\to \mathbb R$ is integrable over the rectangle $A$. Let $B_{1/n}=\{x:\ o(f,x)\geq 1/n\}$. Then $B_{1/n}$ has measure (or, equivalently, content) zero.

The flawed proof is the following.

If $\varepsilon > 0$, let $P$ be a partition of $A$ such that $U(f,P)-L(f,P)<\varepsilon/n$. Let $\cal S%$ be the collection of subrectangles $S$ of $P$ which intersect $B_{1/n}$. Then $\cal S$ is a cover of $B_{1/n}$. Now if $S\in \cal S$, then $M_S(f)-m_S(f)\geq 1/n$. Thus: $$\dfrac{1}{n}\sum _{S\in\cal S}v(S)\leq \sum _{S\in \cal S} [M_S(f)-m_S(f)]\cdot v(S)\leq U(P,f)-L(P,f)\leq\varepsilon /n.$$ Hence $B_{1/n}$ has content zero.

The part I've emphasized in bold character is the wrong one, since it is true only if $B_{1/n}$ intersects $S$ in its interior. The proposed solution:

[...]To compensate for this it suffices to cover the boundaries of all subrectangles of $P$ with a finite collection of rectangles with total volume $<\varepsilon$. These, together with $\cal S$, cover $B_{1/n}$ and have total volume $<2\varepsilon$.

My bad, I don't have a great intuition of $n$-dimensional rectangles, I don't understand what he is suggesting to do. So I ask your help in understanding the last statement or (even better) if you can fix the proof in another way.

My alternative idea was to cover each $S\in\cal S$, with an open rectangle $S'$, such that $v(S'-S)<c\varepsilon $, but has brought me nowhere.

It may be useful to clarify some notations: $$o(f,x)=\lim _{\delta \to 0} \{ \sup _{|t-x|<\delta} f(t)- \inf _{|t-x|<\delta} f(t)\}\\M_S(f)=\sup _{x\in S} f,\quad m_s(f)=\inf _{x\in S} f.$$ $U(P,f)$, and $L$ are Rieman upper and lower sums.

pppqqq
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1 Answers1

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The set $B_{1/n}$ can be covered by those subrectangles whose interior intersect $B_{1/n}$, call this set $\mathcal{S}$, and let $\mathcal{T}$ be the set of all boundaries of subrectangles. Then $\mathcal{S}\cup \mathcal{T}$ is a cover for for $B$, as every subrectangle intersects that intersects $B_{1/n}$ either interests in its interior with $B_{1/n}$ or otherwise the intersection lives inside the boundary of this subrectangle. You can show that $\mathcal{S}$ has small volume by an argument above. It remains to show that $\mathcal{T}$ has small volume, this has two do with two general facts: (i) if $R$ is a rectangle in $\mathbb{R}^{n}$ then $\partial R$ has zero content in $\mathbb{R}$ and (ii) the finite union of sets of content zero is of content zero.

Nicolas Bourbaki
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