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Let $I$ denote a closed (hyper)rectangle in $R^n$ and $f:I \rightarrow \mathbb{R}$ be a bounded function. We use the following notations.

(1) $\nu(f;a)$ is the oscillation of the function $f$ at a point $a \in I$, defined as follows: $$ \nu(f;a) = \lim_{\delta \to 0} \left[ \sup \{ |f(x)-f(y)| : x, y \in B(a;\delta) \cap I \} \right] $$ where $B(a;\delta)$ denotes the open ball of radius $\delta$ centered at $a$.

(2) For a partition $P$ of $I$, $U(f,P)$ and $L(f,P)$ denote the upper and lower Riemann sums of $f$ over $P$. For completeness, the definitions are as follows: $$ U(f,P) = \sum_{R} M_R(f) \cdot V(R), \qquad L(f,P) = \sum_{R} m_R(f) \cdot V(R) $$ where the sums are taken over the set of all subrectangles $R$ formed by P, $V(R)$ denotes the volume of $R$, and $M_R(f) = \sup \{ f(x) : x \in R\}$, $m_R(f) = \inf \{ f(x) : x \in R\}$.

Question:

Suppose there is a number $\epsilon>0$ such that $\nu(f;x) < \epsilon$ for every $x \in I$. Show that there exists a partition $P$ of $I$ such that $U(f,P) - L(f,P) < \epsilon \cdot V(I).$

A have an idea of how to begin the proof, but couldn't proceed after a point. Here is how the argument goes: For every point $x \in I$ choose a rectangular open neighborhood $V_x$. The collection of sets $V_x$ for $x \in I$ is an open cover for $I$, hence it must have a finite subcover $V_1,\ldots,V_m$ by compactness of $I$ in $\mathbb{R}^n$. Denote $V_j \cap I$ by $V'_j$. Then $V'_1,\ldots,V'_m$ is a finite cover for $I$. Since $\nu(f;x) < \epsilon \ \forall x \in V'_j \ \forall j$ I expect we can choose a partition $P$ such that each subrectangle $R$ satisfies $M_R(f) - m_R(f) < \epsilon$ by somehow making use of the rectangles $V'_j$. But I am unable to make this precise since the rectangles $V'_j$ are not closed.

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$\nu(f,a) < \epsilon$ means that there exists $\delta(a) > 0$ such that $\sup \{ \lvert f(x) - f(y) \rvert : x,y \in B(a;\delta(a)) \cap I \} < \epsilon$.

Since $\nu(f,a) < \epsilon$ for all $a \in I$, we conclude that if a subrectangle $R \subset I$ is contained in some $B(a;\delta(a)) \cap I$, then $M_R(f) -m_R(f) < \epsilon$.

$\mathcal{B} = \{ B(a;\delta(a)) \cap I : a \in I \}$ is an open cover of $I$. There exists a Lebesgue number $\rho > 0$ for $\mathcal{B}$. This means that that every subset of $I$ having diameter less than $\rho$ is contained in some member of $\mathcal{B}$.

Now choose a partition $P$ of $I$ such that all elements have diameter less than $\rho$.

Edited:

I add a proof for the existence of a Lebesgue number. I shall do this in a form specialized to your situation.

The diameter of a bounded subset $M \subset \mathbb{R}^n$ is defined as

$$diam(M) = \sup \{ \lvert x - y \rvert : x,y \in M \} .$$

We claim that there exists $\rho > 0$ such that for all $M \subset I$ with $diam(M) < \rho$ there exists $a \in I$ such that $M \subset B(a;\delta(a)) \cap I$.

If our claim would not be true, then we could find a sequence of nonempty subsets $M_n \subset I$ with $diam(M_n) < \frac{1}{n}$ which are contained in no $B(a;\delta(a))$. Choose $x_n \in M_n$. Then $(x_n)$ is a sequence in $I$ and must have a subsequence converging to some $x \in I$ because $I$ is compact. Therefore we can find a sufficiently large $n$ such that both $diam(M_n) < \frac{1}{2}\delta(x)$ and $\lvert x_n - x \rvert < \frac{1}{2}\delta(x)$. Hence for $y \in M_n$ $$\lvert y - x \rvert \le \lvert y - x_n \rvert + \lvert x_n - x \rvert < \delta(x)$$ which means $M_n \subset B(x;\delta(x)$, a contradiction.

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  • $\begingroup$ $B(a,\delta(a)) \cap I$ need not be open, for example if $a$ is a boundary point of $I$. So I don't think $\mathcal{B}$ is an open cover for $I$. Also is there a more 'elementary' way to prove this without using the concept of Lebesgue number (as I am not familiar with it)? $\endgroup$ Oct 12, 2018 at 11:09
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    $\begingroup$ The $B(a;\delta(a)) \cap I$ are in general not open in $\mathbb{R}^n$, but they are open in the subspace $I \subset \mathbb{R}^n$ (remember the subspace topology). Therefore $\mathcal{B}$ is an open cover of the compact metric space $I$. $\endgroup$
    – Paul Frost
    Oct 12, 2018 at 12:04
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    $\begingroup$ An open cover of a space $X$ is a cover with open subsets of $X$. If $X \subset Y$, you can also consider covers of $X$ by open subset $U_\alpha \subset Y$, (i.e. $X \subset \bigcup_\alpha U_\alpha$), but then the $U_\alpha \cap X$ form an open cover of $X$ in the usual sense. $\endgroup$
    – Paul Frost
    Oct 12, 2018 at 12:13
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    $\begingroup$ I do not see a more elementary proof, but perhaps somebody else can help. Lebesgue numbers are a standard tool used for compact metric spaces. You will find it in most textsbooks. If you want, I shall provide a proof for the existence of Lebesgue numbers. $\endgroup$
    – Paul Frost
    Oct 12, 2018 at 12:18
  • $\begingroup$ Thanks for the edit, it was a real help. $\endgroup$ Oct 13, 2018 at 15:14

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