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Let $f:[a,b]\longrightarrow \mathbb{R}$

Let $P=\{a=t_0<t_1<\cdots<t_n=b\}$ be partition of $[a,b]$

$P^*=(P,\xi)$ , $\xi=(\xi_1,\xi_2,\cdots,\xi_n)$ , $t_{i-1}\le\xi_i \le t_i$

We define $\displaystyle \sum (f,P^*)=\sum_{i=1}^{n} f(\xi_i) \cdot (t_i-t_{i-1}) $

$||P||=\underset {1\le i\le n}{\text{max}} (t_i-t_{i-1})$


Definition $1$

We define $\displaystyle \int_a^b f =\lim_{||P||\to0} \sum(f,P^*)$ if the limit exists

$\forall \epsilon >0 , \exists\delta>0: \forall P^* ,||P||<\delta \Longrightarrow \left|\displaystyle \int_a^b f-\sum(f,P^*)\right|<\epsilon$


Definition $2$

Another definition of Riemann integral:

If there is a number $L$ such that:

$\forall \epsilon >0 , \exists P_{\epsilon} $ partition of $[a,b]: \forall P^*=(P,\xi) , P_{\epsilon}\subset P $ finer partition $ \Longrightarrow |L-\sum(f,P^*)|<\epsilon$

then $\displaystyle \int_a^b f=L$


We note that Definition $1$ implies Definition $2$, but what about the converse ?

Any hints would be appreciated.

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  • $\begingroup$ Oh, you wrote "Riemann Stieltjes" and I was misled to believe you wanted to prove it fails in the general case. Yes, for the usual Riemann integral they are equivalent. Maybe you can make this clearer? $\endgroup$ – Pedro Tamaroff Jun 1 '13 at 1:43
  • $\begingroup$ I edited my answer. $\endgroup$ – Pedro Tamaroff Jun 1 '13 at 1:54
  • $\begingroup$ What does $\lim_{\lVert P \rVert\to 0}$ mean, exactly? It is not very easy to define such a thing in a rigorous way. See here $\endgroup$ – Giuseppe Negro Jun 1 '13 at 2:07
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    $\begingroup$ Why does definition 1 imply definition 2? $\endgroup$ – Monolite Feb 23 '15 at 1:05
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This is a sketch presented in Apostol's Mathematical Analysis. Can you fill in the details? There aren't too many left! You need to look at the $S_1,S_2$ carefully.

Let $\int_a^b f(x)dx =I$, $M=\sup\{|f(x)|:x\in[a,b]\}$. Given $\epsilon >0$; choose $P_\epsilon$ such that $U(P_\epsilon,f)<I+\dfrac \epsilon 2$. [Here $U$ is the upper Darboux/Riemann sum of $f$] Let $N$ be the number of points of division in $P_\epsilon$ and let $\delta=\dfrac{\epsilon}{2MN}$. If $\Vert P\Vert <\delta$ put $$U(f,P)=\sum M_kf(x) \Delta_x=S_1+S_2$$ where $S_1$ is the sum over the terms that belong to the subintervals of $P$ that have no points of $P_\epsilon$ and $S_2$ is the remianing sum. Then $$S_1\leq U(P_\epsilon,f)<I+\frac \epsilon 2$$ $$S_2\leq NM\Vert P\Vert <NM\delta=\frac\epsilon 2$$ thus $$U(P,f)<I+\epsilon$$ Analogously, $$L(f,P)>I-\epsilon$$ if $\Vert P\Vert<\delta^\prime$, for a suitable $\delta^\prime$. Thus $$|S(P,f)-I|<\epsilon$$ if $\Vert P\Vert<\min\{\delta,\delta^{\prime}\}$.

NOTATION $U(f,P)$ denotes the upper sum of $f$ w.r.t $P$, that is $$\sum M_k \Delta x_k$$ where $M_i$ is the supremum of $f$ over the $i$-th subinterval of $P$. Analogously, $L(f,P)$ is the lower sum of $f$ w.r.t. $P$. Finally $S(f,P)$ is any Riemann sum of a tagged partition $P$.

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  • $\begingroup$ But in this case $\alpha(t)=t$ , $\forall t \in [a,b]$ $\displaystyle \int_a^b f d \alpha = \int_a^b f$ $\endgroup$ – felipeuni Jun 1 '13 at 0:21
  • $\begingroup$ But the question is whether the Definition $2$ implies Definition $1$ if $\alpha(t)=t$ , $\forall t \in [a,b]$ $\endgroup$ – felipeuni Jun 1 '13 at 1:35
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    $\begingroup$ @PedroTamaroff Which page in Apostol's can I find this? $\endgroup$ – Heisenberg Dec 16 '13 at 8:55
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    $\begingroup$ @Heisenberg this is exercise 7.4 there. $\endgroup$ – leo Jan 19 '14 at 22:42
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    $\begingroup$ Two questions about the very beginning of you proof. 1) You write "choose $P_\varepsilon$ such that", but the definition assume the existence of some $P_\varepsilon$; why are we allowed to choose one? 2) The same definition use the Riemann sum and not the upper Darboux sum. $\endgroup$ – Aaron Lenz May 9 at 17:39
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When proving that a function which is integrable according Definition 2 is also integrable according to Definition 1, given $\epsilon\gt 0$ one is tempted to take $\delta$ as $\Vert P_\epsilon\Vert$. The problem is that there can be partitions $P$ with $\Vert P\Vert \lt \delta$ but $P\not\supseteq P_\epsilon$. So, how will we use Definition 2 then? Let's see.

We say that $f\in \mathscr R[a,b]$ when $f$ is integrable according to Definition 2. We also write $\mathscr P[a,b]$ for the set of partitions of the interval $[a,b]$. As usual $\Vert f\Vert_\infty =\sup\{|f(x)| : x\in [a,b]\}$.

Lemma Let $f:[a,b]\to \Bbb R$ be a bounded function, and $P\in \mathscr P[a,b]$.

  1. If $P_1$ is the partition obtained from $P$ by adding one point then \begin{gather*} L(P_1,f)\leq L(P,f) + 2\Vert f\Vert_\infty \Vert P\Vert\\ U(P_1,f)\geq U(P,f) - 2\Vert f\Vert_\infty \Vert P\Vert \end{gather*}
  2. If $P_m$ is the partition obtained from $P$ by adding $m$ points then \begin{gather*} L(P_m,f)\leq L(P,f) + 2m\Vert f\Vert_\infty \Vert P\Vert\\ U(P_m,f)\geq U(P,f) - 2m\Vert f\Vert_\infty \Vert P\Vert \end{gather*}

Proof.

  1. Let $z\in [a,b]$ be the point to be added. If $P = \{a=x_0,x_1\ldots,x_n=b\}$, there is some $k\in\{1,\ldots,n\}$ so that $z\in [x_{k-1},x_k]$. Thus, letting \begin{gather*} m^\prime = \inf f\left(\left[x_{k-1},z\right]\right)\\ m^{\prime\prime} = \inf f\left(\left[z,x_k\right]\right) \end{gather*} we have \begin{align*} L(P,f) - L(P_1,f) &= m_k(f)\left(x_k - x_{k-1}\right) - m^\prime\left(z - x_{k-1}\right) - m^{\prime\prime}\left(x_k - z\right)\\ &\leq \Vert f\Vert_\infty\left(x_k - x_{k-1}\right) + \Vert f\Vert_\infty\left(z - x_{k-1}\right) + \Vert f\Vert_\infty\left(x_k - z\right)\\ &= 2\Vert f\Vert_\infty\left( x_k - x_{k-1}\right)\\ &\leq 2\Vert f\Vert_\infty\Vert P\Vert. \end{align*} The other inequality is similar.

  2. This follows by adding one point at time and iterating the inequalities in 1.

Now, suppose $f\in \mathscr R[a,b]$. Let $\epsilon\gt 0$. Then we know that $$ \underline{\int_a^b} f(x)\mathrm d x = \int_a^b f(x)\mathrm d x = \overline{\int_a^b} f(x)\mathrm d x, $$ so by properties of sup (and inf), there exist $P_1,P_2\in\mathscr P[a,b]$ such that if $P\in\mathscr P[a,b]$ with $P\supseteq P_1$ ($P\supseteq P_2$) then \begin{gather*} \int_a^b f(x)\mathrm d x - \frac\epsilon 2 \lt L\left(P_1,f\right)\\ \int_a^b f(x)\mathrm d x + \frac\epsilon 2 \gt U\left(P_2,f\right). \end{gather*} Let $$ P=P_1\cup P_2 = \{x_0,\ldots,x_n\} $$ and $\delta = \frac\epsilon {4n\Vert f\Vert_\infty}$. Let $Q$ be a partition of $[a,b]$, with $\Vert Q\Vert\lt \delta$. Then $P\cup Q$ is obtained from $Q$ by adding at most $n-1$ points. Remember that when you add points to a partition, lower sums increases and upper sums decreases, thus \begin{gather*} \int_a^b f(x) \mathrm d x -\frac\epsilon 2 \lt L(P_1,f)\leq L(P\cup Q,f) \leq L(Q,f) + 2(n-1)\Vert f\Vert_\infty\Vert Q\Vert.\\ \int_a^b f(x) \mathrm d x -\frac\epsilon 2 \lt L(Q,f) + 2(n-1)\Vert f\Vert_\infty \frac\epsilon{4n\Vert f\Vert_\infty}\\ \int_a^b f(x) \mathrm d x -\frac\epsilon 2 \lt L(Q,f) + \frac\epsilon 2 \\ \int_a^b f(x) \mathrm d x -\frac\epsilon 2 \lt \sum (Q^\ast,f) + \frac\epsilon 2\\ \int_a^b f(x) \mathrm d x \lt \sum(Q^\ast,f) + \epsilon. \end{gather*} Similarly, using upper sums we get $$\int_a^b f(x) \mathrm d x \gt \sum(Q^\ast,f) - \epsilon.$$ Therefore $$\left| \int_a^b f(x) \mathrm d x - \sum(Q^\ast,f)\right|\lt \epsilon,$$ whenever $\Vert Q\Vert\lt\delta$, that is, $f$ is integrable according to Definition 1.

Notes This approach was found as Exercise 19 in Section 7.1 of the Spanish translation of Introduction to Real Analysis by Robert Bartle and Donald Shertbert. It was the second edition. Once I searched for that exercise in the newest edition. I couldn't fInd it. So Here it is.

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