Will we have convergence to the Riemann integral?

Question

Assume you have K functions $f_1,f_2,\ldots,f_K$, we assume that each of the functions is Riemann-integrable, and that their product is Riemann-integrable.

We also assume that we have a sequence of partitions of the set $[a,b]$ that is a function of $m$, where the norm of the partition goes to zero as $m$ goes to infinity. $N_m$ is the number of subintervals for each $m$.

For each $j_m \in \{0,\ldots,N_m-1\}$ and $i \in \{1,\ldots,K\}$ we assume that $c_{j_m}^{(i)}$ is in the corresponding subinterval. We define $\Delta x_{j_m}$ as the length of the subinterval.

Will then the sequence in $m$ defined by:

$\sum\limits_{j_m=0}^{N_m-1}f_1\left(c_{j_m}^{(1)}\right)f_2\left(c_{j_m}^{(2)}\right)\cdots f_K\left(c_{j_m}^{(K)}\right)\Delta x_{j_m}$

converge to the corresponding Riemann-integral?

The problem here is that is not a Riemann-sum since $c_{j_m}^{(1)}, c_{j_m}^{(2)},\ldots, c_{j_m}^{(K)}$ might differ(but they are still in the same subinterval).

Answer 1

The idea is pretty clear, but we just have to write a lot, and use a lot of notation.

Definitions. A partition of $[a,b]$ is a finite sequence $\pi = (a_j)_{j=0}^J$ with $a=a_0 < a_1 < \cdots < a_J = b$.
The norm of $\pi$ is $\|\pi\| = \max_{1 \le j \le J} |a_j-a_{j-1}|$.
A tagged partition of $[a,b]$ is a pair $(\pi,\gamma)$ where $\pi = (a_j)_{j=0}^J$ is a partition of $[a,b]$ and the finite squence $\gamma = (c_j)_{j=1}^J$ satisfies $a_{j-1} \le c_j \le a_j$ for all $j$.

Let $f : [a,b] \to \mathbb R$. Let $\pi = (a_j)_{j=0}^J$ be a partition of $[a,b]$.
The upper sum is $$ U(f,\pi) = \sum_{j=1}^J M_j(f)\;\cdot(a_j-a_{j-1})\qquad\text{where}\qquad M_j(f) = \sup \{f(x) : a_{j-1} \le x \le a_j\} . $$
The lower sum is $$ L(f,\pi) = \sum_{j=1}^J m_j(f)\;\cdot(a_j-a_{j-1})\qquad\text{where}\qquad m_j(f) = \inf \{f(x) : a_{j-1} \le x \le a_j\} . $$

Let $(\pi,\gamma)$ be a tagged partition. The Riemann sum is $$ R(f,\pi,\gamma) = \sum_{j=1}^J f(c_j)\cdot (a_j-a_{j-1}) $$

We say that $f$ is Riemann integrable, and its integral is $V$ iff
for every $\epsilon > 0$ there exists $\delta > 0$ such that for every tagged partition $(\pi,\gamma)$ of $[a,b]$, if $\|\pi\| < \delta$, then $$ \big |R(f,\pi,\gamma) - V \big| < \epsilon . $$

Alternatively (from a Cauchy criterion), $f$ is Riemann integrable if and only if: for every $\epsilon > 0$ there exists $\delta > 0$ such that for every partition $\pi$ of $[a,b]$, if $\|\pi\| < \delta$ then $$ U(f,\pi) - L(f,\pi) < \epsilon . $$ If so, then the integral $V$ is the unique number with $L(f,\pi) \le V \le U(f,\pi)$ for all partitions $\pi$ of $[a,b]$.

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Theorem A
Let $f_1,f_1,\dots, f_K$ be functions on $[a,b]$ such that $f_1, f_2,\dots, f_K$ and the product $F = f_1 f_2\cdots f_K$ are all Riemann integrable.

Claim: for any $\epsilon > 0$ there exists $\delta > 0$ such that: if $\pi$ is a partition of $[a,b]$ and $\|\pi\| < \delta$ and $\gamma^{(k)}=(c_j^{(k)})_{j=1}^J$ , $1 \le k \le K$, are $K$ choices of tags so that for all $k$, the pair $(\pi,\gamma^{(k)})$ is a tagged partition of $[a,b]$, then $$ \left|U(F,\pi) - \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)})\cdot (a_j-a_{j-1})\right| < \epsilon \tag1$$ and $$ \left|\sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)})\cdot (a_j-a_{j-1})-L(F,\pi)\right| < \epsilon \tag2$$ and $$ U(F,\pi) - L(F,\pi) < \epsilon. \tag3$$

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Case 1: All $f_k \ge 0$.
So, $\epsilon > 0$ is given. Since all the functions $f_k$ are Riemann integrable, they are bounded. There is a single bound $A > 0$ with $$ |f_k(x)|\le A \qquad\text{for}\qquad k=1,\dots,K,\qquad a\le x \le b . $$

Let $$\epsilon' = \frac{\epsilon}{K A^{K-1}}.$$ Since $f_1,\dots,f_K$ are all Riemann integrable, there is $\delta > 0$ so that for any partition $\pi = (a_j)_{j=0}^J$ of $[a,b]$, if $\|\pi\| < \delta$, then $$ U(f_k,\pi) - L(f_k,\pi) < \epsilon'\qquad\text{for } k=1,\dots,K . $$ Let $(\pi, \gamma^{(k)})$ be tagged partitions as above. Assume $\|\pi\| < \delta$. Write $\widetilde{M}_j = M_j(f_1)M_j(f_2)\cdots M_j(f_K)$ and $\widetilde{m}_j = m_j(f_1)m_j(f_2)\cdots m_j(f_K)$.

Write $$ \widetilde{U} = \sum_{j=1}^J \widetilde{M}_j\cdot(a_j-a_{j-1}),\qquad \widetilde{L} = \sum_{j=1}^J \widetilde{m}_j\cdot(a_j-a_{j-1}) $$ Claim 1: $\widetilde{U}-\widetilde{L} < \epsilon$

Claim 2: All three of $$ U(F,\pi), \quad L(F,\pi), \quad \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_J(c_j^{(K)})\cdot (a_j-a_{j-1}) $$ are between $\widetilde{U}$ and $\widetilde{L}$.
Consequently, we obtain $(1), (2), (3)$, as required.

Proof of Claim 1.
The difference $\widetilde{U}-\widetilde{L}$ is written as a sum of $L$ terms. First, fix $j$ between $1$ and $J$. Then \begin{align} \widetilde{M}_j &= M_j(f_1)M_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ & \ge m_j(f_1)M_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ & \ge m_j(f_1)m_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ & \ge \cdots \\ & \ge m_j(f_1)m_j(f_2)\cdots m_n(f_K) = \widetilde{m}_j \end{align} From each row to the next, one $M_j$ changes to an $m_j$. Then \begin{align} \widetilde{M}_j - \widetilde{m}_j &= (M_j(f_1)-m_j(f_1)) M_j(f_2)M_j(f_3)\cdots M_j(f_K) \\ &+ m_j(f_1)(M_n(f_2)-m_j(f_2)))M_j(f_3)\cdots M_j(f_K) \\ &+ \cdots \\&+ m_j(f_1) m_j(f_2)\cdots m_j(f_{K-1})(M_j(f_K)-m_j(f_K)) \\ &\le A^{K-1}\sum_{k=1}^K (M_j(f_k) - m_j(f_k)) \end{align} Now multiply by $a_j-a_{j-1}$ and sum on $j$. The left side is $$ \sum_{j=1}^J (\widetilde{M}_j - \widetilde{m}_j)\cdot(a_j-a_{j-1}) =\widetilde{U} - \widetilde{L}. $$ The right side is \begin{align} A^{K-1}\sum_{k=1}^K &\sum_{j=1}^J (M_j(f_k) - m_j(f_k))\cdot(a_j-a_{j-1}) = A^{K-1}\sum_{k=1}^K \big(U(f_k,\pi) - L(f_k,\pi)\big) \\ & < A^{K-1} K \epsilon' =\epsilon \end{align} This completes the proof of Claim 1.

Proof of Claim 2.
To prove $\widetilde{U} \ge U(F,\pi) \ge \widetilde{L}$ note that $\widetilde{M}_j \ge M_j(F) \ge \widetilde{m}_j$. To prove $\widetilde{U} \ge L(F,\pi) \ge \widetilde{L}$ note that $\widetilde{M}_j \ge m_j(F) \ge \widetilde{m}_j$. To prove $$ \widetilde{U} \ge \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_J(c_j^{(K)})\cdot (a_j-a_{j-1}) \ge \widetilde{L} $$ note \begin{align} \widetilde{M}_j &= M_j(f_1)M_j(f_2)\cdots M_j(f_K) \\ &\ge f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)}) \\ &\ge m_j(f_1)m_j(f_2)\cdots m_j(f_K) = \widetilde{m}_j \end{align} multiply by $a_j-a_{j-1}$ and sum on $j$.

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Case 2. arbitrary signs.
Write $f_k = f_k^+ - f_k^-$ in positive and negative parts. Then $f_1 f_2\cdots f_k$ is a linear combination of $2^K$ such sums with nonnegative functions. Both $$ \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)})\cdot (a_j-a_{j-1}) \qquad\text{and}\qquad \int_a^b f_1(x)f_2(x)\cdots f_K(x)\;dx $$ are linear combinations of $2^K$ terms of the same form, but involving only nonnegative functions. Theorem A for $f_1f_2\cdots f_k$ follows from those $2^k$ cases of Theorem A for nonnegative functions.

Answer 2

The proof is quite straightforward, although it requires some wording. Before the proof, let us clear out some definitions. Once we set up the notations, the actual proof will be just a couple of lines. $ \newcommand{\ttbr}[1]{\mathtt{[}{#1}\mathtt{]}} \newcommand{\ttP}{\mathtt{P}} \newcommand{\ttc}{\mathtt{c}} $

Definitions.

• A partition of $[a, b]$ is a finite subset of $[a, b]$ containing both $a$ and $b$. If $\ttP$ is a partition with $N+1$ elements, we often write

  $$\ttP = \{\ttP\ttbr{0} < \ttP\ttbr{1} < \cdots < \ttP\ttbr{N}\}.$$

  For the convenience, we will also write

  $$ \#\ttP := N, \qquad \Delta\ttP\ttbr{j} = [\ttP\ttbr{j-1}, \ttP\ttbr{j}], \qquad \|\ttP\| = \max_{1\leq j \leq \#\ttP} |\Delta\ttP\ttbr{j}| $$

  for the length, $j$-th sub-interval, and mesh size of $\ttP$, respectively.

• Let $\ttP$ be a partition. Then $\ttc = (\ttc\ttbr{1}, \cdots, \ttc\ttbr{\#\ttP})$ is called a tag of $\ttP$ whenever

  $$\ttc\ttbr{j} \in \Delta\ttP\ttbr{j}, \qquad \forall i \in \{ 1, \cdots, \#\ttP \}. $$

• Let $\ttP$ be a partition of $[a, b]$ with a tag $\ttc$. For each bounded $f : [a, b] \to \mathbb{R}$, we define the upper/lower Darboux sums and the Riemann sum by

  $$ \begin{gathered} U(f; \ttP) = \sum_{j=1}^{\#\ttP} \bigg( \sup_{x \in \Delta\ttP\ttbr{j}} f(x) \bigg) \, |\Delta\ttP\ttbr{j}|, \\ L(f; \ttP) = \sum_{j=1}^{\#\ttP} \bigg( \inf_{x \in \Delta\ttP\ttbr{j}} f(x) \bigg) \, |\Delta\ttP\ttbr{j}|, \\ R(f; \ttP; \ttc) = \sum_{j=1}^{\#\ttP} f(\ttc\ttbr{j}) \, |\Delta\ttP\ttbr{j}|. \end{gathered} $$

• For $f : [a, b] \to \mathbb{R}$, its supremum norm is defined by $\|f\| = \sup\{|f(x)| : x \in [a, b]\}$.

Now we return to the proof. Recall the following equivalence:

Proposition. Let $f : [a, b] \to \mathbb{R}$ be bounded and $L \in \mathbb{R}$. Then the followings are equivalent:

1. $f$ is Riemann-integrable with the value $L$, i.e., for any $\epsilon > 0$ there exists $\delta > 0$ such that whenever $\ttP$ is a partition of $[a, b]$ with $\|\ttP\| < \delta$ and $\ttc$ is a tag of $\ttP$, we have $|R(f; \ttP; \ttc) - L| < \epsilon$.

2. $f$ is Darboux-integrable with the value $L$, i.e., for any $\epsilon > 0$ there exists $\delta > 0$ such that whenever $\ttP$ is a partition of $[a, b]$ with $\|\ttP\|<\delta$, we have $|U(f;\ttP)-L| < \epsilon$ and $|L(f;\ttP) - L| < \epsilon$.

In light of this, we will interchangeably use any of these statements for Riemann-integrability. Next, we make the following simple observation:

Lemma. Let $f : [a, b] \to \mathbb{R}$ be bounded. Then for any partition $\ttP$ of $[a, b]$ and for any tags $\ttc, \ttc'$ of $\ttP$, we have

$$ \sum_{j=1}^{\#\ttP} | f(\ttc\ttbr{j}) - f(\ttc'\ttbr{j}) | \, |\Delta\ttP\ttbr{j}| \leq U(f; \ttP) - L(f; \ttP). $$

Proof. This is an immediate consequence of the inequality $|f(x) - f(y)| \leq \sup_A f - \inf_A f$ which holds for any $f : A \to \mathbb{R}$ and for any $ x, y \in A $. ////

Now we are ready to prove the statement. Let $f_1, \cdots, f_K$ be bounded on $[a, b]$. Let $\ttP$ be a partition of $[a, b]$ and $\ttc_1, \cdots, \ttc_K$ be tags of $\ttP$. For brevity, let us write

$$ S(\ttc_1, \cdots, \ttc_K) = \sum_{j=1}^{\#\ttP} f_1(\ttc_1\ttbr{j}) \cdots f_K(\ttc_K\ttbr{j}) \, |\Delta\ttP\ttbr{j}|. $$

In particular, for any tag $\ttc$ of $\ttP$, we know that $S(\ttc, \cdots, \ttc) = R(f_1 \cdots f_K; \ttP; \ttc)$ is the ordinary Riemann sum for the product $f_1 \cdots f_K$. Now our goal is to estimate the difference between $S(\ttc_1, \cdots, \ttc_K)$ and $R(f_1 \cdots f_K; \ttP; \ttc)$. By triangle inequality,

\begin{align*} &\left| S(\ttc_1, \cdots, \ttc_K) - R(f_1 \cdots f_K; \ttP; \ttc) \right| \\ &\quad \leq \sum_{k=1}^{K} \big| S(\ttc_1, \cdots, \ttc_k, \ttc, \cdots, \ttc) - S(\ttc_1, \cdots, \ttc_{k-1}, \ttc, \cdots, \ttc) \big| \\ &\quad \leq \sum_{k=1}^{K} \sum_{j=1}^{\#\ttP} \|f_1\| \cdots \|f_{k-1} \| \big| f_k(\ttc_k\ttbr{j}) - f_k(\ttc\ttbr{j}) \big| \|f_{k+1}\| \cdots \|f_K\| \, |\Delta\ttP\ttbr{j}| \\ &\quad \leq \sum_{k=1}^{K} \left( \prod_{i \neq k} \|f_i\| \right) \left( U(f_k; \ttP) - L(f_k; \ttP) \right). \end{align*}

The above inequality holds for any bounded $f_1, \cdots, f_K$ on $[a, b]$. From now on, we also assume that $f_1, \cdots, f_K$ are Riemann-integrable. Then this bound converges to $0$ as $\|\ttP\| \to 0$. Since the Riemann sum $R(f_1 \cdots f_K; \ttP; \ttc)$ converges to the Riemann integral $\int_{a}^{b} f_1(x)\cdots f_K(x) \, \mathrm{d}x$ as $\|\ttP\| \to 0$, it follows that

$$ \lim_{\|\ttP\| \to 0} \sum_{j=1}^{\#\ttP} f_1(\ttc_1\ttbr{j}) \cdots f_K(\ttc_K\ttbr{j}) \, |\Delta\ttP\ttbr{j}| = \int_{a}^{b} f_1(x)\cdots f_K(x) \, \mathrm{d}x $$

as desired. ////