Completing a proof on integrability: $f$ is Riemann integrable under the mesh definition iff $\sup L_\Gamma = \inf U_\Gamma$

Question

$ \newcommand{\II}{\mathcal{I}} \newcommand{\R}{\mathbb{R}} \newcommand{\ve}{\varepsilon} \newcommand{\d}{\delta} \newcommand{\g}{\Gamma} \newcommand{\norm}[1]{\left \Vert #1 \right \Vert} \newcommand{\volume}{\mathrm{volume}} \newcommand{\gs}{{\g_\ast}} $Definitions:

Considering we're working in $\R^n$, and there are a lot of competing/equivalent definitions for Riemann integrability (as I learned the hard way in trying to find some help here and elsewhere), I want to get all of these out of the way first, so we're on the same page. These are the definitions I'm working with, in their essence taken from Measure & Integral: An Introduction to Real Analysis by Richard Wheeden and Antoni Zygmund.

• Riemann Integrable: Let $f : \mathcal I \subseteq \R^n \to \R$ be bounded. We say $f$ is Riemann integrable on $\II$ if $$ \lim_{\norm \g \to 0} R_\g $$ exists. In particular, we call this value $L$, and say $L = \int_\II f(x) \, \mathrm{d}x$.

• Formalization: Formally, the above is equivalent to

$$(\forall \ve >0)(\exists \d > 0)(\forall \g : \norm \g < \d)(|R_\g - L| < \ve)$$

• Boxes: $\II$ denotes a box. You can write it a couple of different ways: for given $a_i,b_i$, $$ \II := \Big\{ (x_1,\cdots,x_n) \in \R^n \; \Big| \; x_i \in [a_i,b_i] \Big\} = \prod_{i=1}^n [a_i,b_i] $$

• Volume of Boxes: The volume of a box is given in the usual, expected way: $$ \volume(\II) = \prod_{i=1}^n (b_i - a_i) $$

• Partitions: Let $\g$ denotes a partition of $\II$ into finitely many nonoverlapping boxes $\II_k$. (That is, $\mathrm{int}(\II_i) \cap \mathrm{int}(\II_j) = \varnothing$ for all $i \ne j$, and $\bigcup \II_k = \II$.)

• Norm/Mesh of Partitions: The mesh or norm of a partition is defined to be $$ \norm \g := \max_i \mathrm{diam}(\II_i) $$

• Diameter of Set: The diameter of a subset of a metric space is the supremum of possible distances between points. In $\R^n$ under the $2$-norm,

$$\mathrm{diam}(\II) := \sup_{x,y \in \II} \norm{x - y}_2$$

• Tags: We take representative points (or tags) $\xi_i \in \II_i$ for each $\II$.

• Riemann Sum: We define the Riemann sum to be (implicitly dependent on the partitioning and tags) $$ R_\g := \sum_i f(\xi_i) \cdot \volume(\II_i) $$

• Upper & Lower Sums: We define the upper and lower sums as so: $$ U_\g := \sum_i \sup_{x \in \II_i} f(x) \cdot \volume(\II_i) \qquad L_\g := \sum_i \inf_{x \in \II_i} f(x) \cdot \volume(\II_i) $$ Trivially, $L_\g \le R_\g \le U_\g$ for any choice of tags $\xi_i$.

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My Goal: I would like to show that $$ \text{$f$ is Riemann integrable} \iff \inf_\g U_\g = \sup_\g L_\g $$ More specifically, I would like to show the backwards direction; I've already proven the forwards I believe. I'd also in particular like as direct a proof as possible (rather than circumnavigating through the other definitions of integrability).

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My Attempt So Far:

So far, I feel like I've hit on the right idea, with a minor snag in the way.

So, let $\ve > 0$, and let $$ \sup_\g L_\g = \inf_\g U_\g =: L $$ From the definition of supremum/infimum, we may find two partitions $\g_1,\g_2$ such that $$ 0 \le \sup_\g L_\g - L_{\g_1} < \frac \ve 2 \qquad 0 \le U_{\g_2} - \inf_\g U_\g < \frac \ve 2 $$ Form their common refinement $\gs$ of $\g_1$ and $\g_2$. (This would just be their union in the $\R^1$ case where we can equivalently use points as our partition, but not the case in general.) Then we easily see that $$ 0 \le \sup_\g L_\g - L_{\gs} < \frac \ve 2 \qquad 0 \le U_{\gs} - \inf_\g U_\g < \frac \ve 2 $$ Recalling the assignment of $L$, we have $$ 0 \le L - L_{\gs} < \frac \ve 2 \qquad 0 \le U_{\gs} - L < \frac \ve 2 $$

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Where I Get Lost:

This is where I get lost. My end goal is to ultimately show $|R_\g - L| < \ve$ (independent of the choice of $\xi_i$ namely). At this point, I feel like I'm meant to use the inequality $L_\g \le R_\g \le U_\g$, but I'm not certain that it would be valid. If it were viable, then I would intend to have $$ 0 \le L - R_{\gs} < \frac \ve 2 \qquad 0 \le R_{\gs} - L < \frac \ve 2 $$ but would it not be possible for $R_\gs$ to fall on one side of $L$ or the other, and thus outside of the bound of $0$? (Though on the other hand, would that even matter? Is the bound of $0$ important for this?) From a situation like this, I would then like to claim the conclusion.

Another worrying part is that I have no particular $\d$ brought up. Maybe I could just arbitrarily set a partition $\g_0$ such that $\norm{\g_1},\norm{\g_2} < \norm{\g_0} = \d$, but that doesn't really feel right.

Can anyone give me an idea as to how to complete this proof, even if just a nudge in the right direction? Or am I completely off-base? Thanks for any insights you can give!

Answer 1

$ \newcommand{\II}{\mathcal{I}} \newcommand{\R}{\Bbb{R}} \newcommand{\g}{\Gamma} \newcommand{\vol}{\operatorname{volume}} \newcommand{\gs}{{\Gamma_\ast}} $

If $U \subseteq V$, then $$\inf_V f \le \inf_U f \le \sup_U f \le \sup_V f$$

Since $\gs$ is a refinement of $\g_1$, every box $\II_1 \in \g_1$ is the union of a collection $\mathscr B(\II_1)$ of boxes in $\gs$, and $$\vol(\II_1) = \sum_{\II_*\in \mathscr B(\II_1)} \vol(\II_*)$$

So $$\begin{align}L_{\g_1} &= \sum_{\II_1\in \g_1} \inf_{\II_1}f\cdot\vol(\II_1)\\ &= \sum_{\II_1\in \g_1} \sum_{\II_*\in \mathscr B(\II_1)} \inf_{\II_1}f\cdot\vol(\II_*)\\ &\le \sum_{\II_1\in \g_1}\sum_{\II_*\in \mathscr B(\II_1)} \inf_{\II_*}f\cdot\vol(\II_*)\\ &= \sum_{\II_*\in \gs}\inf_{\II_*}f\cdot\vol(\II_*)\\ &=L_{\gs}\end{align}$$

Similar remarks hold for the upper sums and for $\g_2$

Answer 2

Let $M = \sup_{x \in \mathcal{I}}|f(x)|$ and define the "width" of $\mathcal{I}$ to be W = $\sup \{b_i-a_i : 1 \leqslant i \leqslant n\}$. The objective is to show that there exists $\delta >0$ such for any partition $\Gamma$ with $\|\Gamma\| < \delta$, we have

$$ \tag{*}\sup_{\Gamma}L_\Gamma - \varepsilon < L _\Gamma \leqslant U_{\Gamma} < \inf_{\Gamma}U_\Gamma + \varepsilon,$$

Focusing first on the right-hand inequality, you have shown for any $\epsilon > 0$ there is a partition $\Gamma_2$ such that

$$U_{\Gamma_2} < \inf_{\Gamma}U_\Gamma + \frac{\varepsilon}{2}$$

We will use the following

Lemma. Let $\Gamma = (P_1,\ldots, P_n)$ be a partition of $\mathcal{I}$ where $P_i$ is a partition of the interval $[a_i,b_i]$ and $\|P \| < \delta $. Suppose $\Gamma'$ is a refinement of $\Gamma$ obtained by adding a single point $c$ in the subinterval $(x_{i,j-1},x_{i,j})$ to the "edge" partition $P_i$. Then we have

$$U_\Gamma < U_{\Gamma'} + 2M W^{n-1}\delta$$

It follows easily by induction that if $q$ such points are added to edge partitions, we have $$U_\Gamma < U_{\Gamma'} + 2Mq W^{n-1}\delta$$

Now take $\delta = \frac{\varepsilon}{4MN\delta W^{n-1}}$ where $N$ is the total number of interior points of the edge partitions of $\Gamma_2$. Let $\Gamma$ be any partition with $\|\Gamma\| < \delta$ and form the common refinment $\Gamma' = \Gamma \cup \Gamma_2$. Since $\Gamma'$ is a refinement of $\Gamma_2$ we have immediately that $U_{\Gamma'} \leqslant U_{\Gamma_2}$. Also, in forming $\Gamma'$ at most $N$ points are added to edge partitions of $\Gamma$ and from the Lemma it follows that

$$U_\Gamma < U_{\Gamma'} + 2MNW^{n-1} \delta = U_{\Gamma'} + 2MNW^{n-1} \cdot \frac{\varepsilon}{4MN\delta W^{n-1}}= U_{\Gamma'} + \frac{\varepsilon}{2} \\< U_{\Gamma_2} + \frac{\varepsilon}{2} < \inf_\Gamma U_\Gamma + \epsilon$$

A similar Lemma and argument can be used to prove that the left-hand inequality in (*) holds.

Proof of Lemma.

Consider the subrectangle $R_{jpq} =S_p \times [x_{i,j-1}, x_{i,j}] \times T_q$ of $\Gamma$, where $S_p \subset \mathbb{R}^{i-1}$ and $T_q \subset \mathbb{R}^{n-i}$. If we add the single point $c \in (a_{i,j-1}, a_{i,j})$ to $P_i$ in forming the refinement $\Gamma'$, then the contribution to $U_\Gamma - U_{\Gamma'}$ from this subrectangle is

$$\sup_{x \in [x_{i,j-1}, c]}f(x)\cdot vol(S_p)vol(T_q)(c- x_{i,j-1})+ \sup_{x \in [c, x_{i,j}]}f(x)\cdot vol(S_p)vol(T_q)(x_{i,j}-c) \\- \sup_{x \in [x_{i,j-1}, x_{i,j}]}f(x)\cdot vol(S_p)vol(T_q)(x_{i,j}-x_{i,j-1}),$$

and, leaving the details to you, we can show by summing over indexes $p$ and $q$,

$$U_\Gamma - U_{\Gamma'} < 2MW^{n-1}\delta$$