Let $Q\subseteq\mathbb{R}^n$ be a rectangle $f:Q\to\mathbb{R}$ be a bounded function. Then for any $\varepsilon>0$ there exists a $\delta>0$ such that $U(f;P)\le \overline{\int}_Qf+\varepsilon$ for any partition $P$ of $Q$ with width (the maximum length of the intervals composing each subrectangle determined by $P$) lesser than $\delta$.
I'm trying to generalize a step from the case $n=1$:
Let $f(x)\ge 0~\forall x\in Q$. Given $\varepsilon>0$ there exists $P_0\in\Pi_Q$ such that $U(f;P_0) < \overline{\int}_Q f + \varepsilon/2$, because the upper integral is the infimum of the upper sums. Let $(R_i)_{1\leq i \leq k}$ be the family of the $k$ rectangles of $P_0$. Choose a $\delta$ such that $0<\delta<{\varepsilon/2kM}$, and take a partition $P$ of $Q$ with width lesser than $\delta$. Denote by $R_\alpha$ the rectangles of $P$ that lie in some $R_i$ of $P_0$, and by $R_\beta$ the remaining rectangles of $P$.
From here, if $n=1$ we could conclude that there are at most $k$ of the rectangles $R_\beta$ because each one of these should have, in their interior, a point of the partition $P_0=\{t_0,\dots t_k\}$ of the closed interval $Q$ (and their interiors are disjoint). And with this we could bound the upper sum the way we want.
But in the general case I can't find a way to do it. I tried considering the border $\partial R_i$ of each rectangle of $P$, and indeed each $R_\beta$ must have points of them in their interior, but the borders infinite sets so I can't find a bound for the number of rectangles $R_\beta$. Any ideas?
Best Answer
Define the width of $Q = [a_1,b_1] \times [a_2,b_2] \times \ldots \times [a_n,b_n]$ to be $W_Q = \max_{1\leqslant j \leqslant n}(b_j - a_j)$. Let $N$ be the number of subrectangles in the partition $P_0$ and take $\delta = \epsilon/(4MN W_Q^{n-1})$ where $|f(x)| \leqslant M$ for all $x \in Q$.
If $\|P \| \leqslant \delta$ and $P'= P \cup P_0$ is the common refinement, then
$$\tag{*}U(f,P) \leqslant U(f,P') + N \cdot 2M \cdot W_Q^{n-1} \cdot \delta \leqslant U(f,P') + \frac{\epsilon}{2}$$
We have $U(f,P') \leqslant U(f,P_0)$ since $P'$ is a refinement of $P_0$, and it follows that
$$U(f,P) \leqslant U(f,P_0) + \frac{\epsilon}{2} \leqslant \overline{\int_Q} f + \epsilon$$
To understand the bound, the partition $P'$ has at most (largely overestimating) $N$ more subintervals of $[a_1,b_1]$ than the partition $P$ and the width of these subintervals is bounded by $\delta$. Each of these extra subintervals $[\alpha_{1j},\beta_{1j}]$ is the edge of multiple rectangles in $P'$ extending in the other dimensions. One such rectangle contributes to the difference $U(f,P) - U(f,P’)$ by no more than the maximum oscillation $2M$ times the volume of the rectangle. The total volume of the rectangles with edge $[\alpha_{1j},\beta_{1j}]$ is conservatively bounded above by $W_Q^{n-1} \cdot \delta$.
Elaboration on first inequality in (*)
Consider a slice $\mathcal{S}$ of $P$-rectangles $R_1,\ldots,R_m$ which can be written as $R_j = [\alpha_{1j},\beta_{1j}]\times T_j$ where the intervals $[\alpha_{1j},\beta_{1j}]$ form a partition of $[a_1,b_1]$ and the $T_j$ are $(n-1)$-dimensional rectangles. Each rectangle $R_j$ is a union of rectangles $R_{jk}\subset R_j$ from the refined partition $P'$.
Let $M_j = \sup_{x \in R_j}f(x)$ and $M_{jk} = \sup_{x \in R_{jk}} f(x)$. Since $|f(x)| \leqslant M$, we have $M_j < M_{jk} + 2M$ (although there is always one rectangle $R_{jk}$ with $M_{jk} = M_j$).
With $R_{jk} = [\alpha_{1,jk}, \beta_{1,jk}] \times T_{jk}$ we have $vol(R_{jk}) \leqslant \delta \,vol(T_{jk})$. In forming the refinement $P'$ by merging $P_0$ and $P$ we create no more than $N$ rectangles in $P'$ where the supremum of $f$ does not coincide with the containing rectangle in $P$.
Thus,
$$\sum_{R_j \in \mathcal{S}}M_j \, vol(R_j) \leqslant \sum_{R_j \in \mathcal{S}}\sum_{R_{jk} \subset R_j} M_{jk} \, vol(R_{jk}) +N \cdot 2M \cdot \delta \cdot \max_{R_{jk} \subset R_j \in \mathcal{S}} vol(T_{jk})$$
Summing over all slices $\mathcal{S}$ of $Q$ we recover the upper sums
$$U(f,P)= \sum_{\mathcal{S}}\sum_{R_j \in \mathcal{S}}M_j \, vol(R_j), \,\, U(f,P') = \sum_{\mathcal{S}}\sum_{R_j \in \mathcal{S}}\sum_{R_{jk} \subset R_j} M_{jk}\, vol(R_{jk}), $$
and obtain the inequality
$$U(f,P) \leqslant U(f,P') + N \cdot 2M \cdot \delta \cdot \sum_{\mathcal{S}}\max_{R_{jk} \subset R_j \in \mathcal{S}} vol(T_{jk}) \\ \leqslant U(f,P') + N \cdot 2M \cdot \delta \cdot W_Q^{n-1} $$