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The definite integral of a continuous function can be defined using the bounded monotone sequence property: see Osgood's Functions of Real Variables, p.110.

(link to full book) (screenshots: page 110, page 111, page 112)

Please can you give some other reference of the same type (or, better, a link or a book where the matter is approached in the same manner) ?

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That seems like the standard definition, not less known. – genepeer May 19 '13 at 11:58
@genepeer I meant not the value but the way to obtain it. Can you explain what do you mean by standard definition ? – Tony Piccolo May 19 '13 at 13:05
This should help: Darboux Integral – genepeer May 19 '13 at 13:33
Was the link I gave not a reference? Jeez, sorry I wasted your time. – genepeer May 20 '13 at 14:11
@Tony: From the US copyright office: "The 1961 Report of the Register of Copyrights on the General Revision of the U.S. Copyright Law cites examples of activities that courts have regarded as fair use: "quotation of excerpts in a review or criticism for purposes of illustration or comment; quotation of short passages in a scholarly or technical work, for illustration or clarification of the author’s observations..." – Zev Chonoles May 21 '13 at 8:32

One can define a function $f:[a,b]\to\Bbb R$ to be regulated if there exists a sequence $\{s_n\}$ of step functions such that $s_n\to f$ uniformly. Recall that if $s$ is a step function with constants $c_i$ and intervals of partition $[x_{i-1},x_i]$ for say $i=1,\dots, r$ we define its integral as over $[a,b]$ as $$\int_a^b s=\sum_{i=1}^r c_i\Delta x_i$$

One can prove that if $s_n,t_n$ are two sequences of step functions that converge uniformly to $f$ then $$\lim \int_a^b s_n=\lim \int_a^b t_n$$ by first proving that for each $\epsilon >0$ there exists $N$ such that $n\geq N$ gives $$|s_n(x)-t_n(x)|<\epsilon$$ We then are led to define the integral of a regulated function as $$\int_a^b f=\lim \int_a^b s_n$$ where $\{s_n\}$ is any sequence of step functions that converge to $f$ uniformly over $[a,b]$.

It is a nice task to show that continuous functions over closed and bounded intervals are indeed regulated in the sense of the definition above.


Let $f:[a,b]\to \Bbb R$ be continuous, $\epsilon >0$be given. Let $P_\epsilon(y)$ mean $-$ for each $\epsilon >0$ and each $y\in[a,b]$ $-$ that there exists a step function $s$ such that $|s-f|<\epsilon$ for $x\in[a,y]$. Define $A=\{y\in[a,b]:P_\epsilon(y)\}$

Then $a\in A$ (trivially) and $A$ is bounded above by $b$, so $\sup A=\alpha$ exists.

$(i)$ First, prove $\alpha=b$ as follows. We know $\alpha\leq b$, since $b$ is an upper bound. Suppose that $\alpha <b$. Then there exists a step $s$ such that approximates $f$ within $\epsilon$ over $[a,\alpha]$. Use that $f$ is continuous at $x=\alpha$ to obtain a $\delta>0$ such that $\alpha+\delta\in A$.

$(ii)$ To show $b\in A$, use again that $f$ is continuous, but now at $x=b$.

It is proven that for each $\epsilon >0$ there exists said step function over $[a,b]$. Now take $\epsilon =1,\frac 1 2,\frac 13,\dots,\frac 1n,\dots $ to obtain said sequence. $\blacktriangle$.

ADD In the above, we needn't proceed by contradiction. What we really are doing is proving that for each $\lambda <b$, there exists a $\delta >0$ such that $\lambda+\delta\in A$, thus showing that $\sup A\geq b$, which gives that $\sup A=b$.

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The technical term is regulated function – Martin May 21 '13 at 11:01
@Martin Thank you. – Pedro Tamaroff May 21 '13 at 11:56
The argument you give for the existence of a step function such that $\lVert s_n - f\rVert_\infty \lt \varepsilon$ is neat. Another approach would be to exploit that $f$ is uniformly continuous on $[a,b]$ so that we can divide $[a,b]$ in $2^{n}$ intervals $I_k$ of length $2^{-n}(b-a) \lt \delta$ with $|f(x) - f(y)| \lt \varepsilon/2$ for all $x,y \in I_k$ with appropriately chosen $\delta$. Or one could directly establish the stronger fact that $f$ is regulated iff right and left limits of $f$ exist for all $x \in [a,b]$. – Martin May 21 '13 at 12:12
@Martin Thank you yet again! I think the proof is the most elementary of them all: it "merely" uses the axiom of completeness. =D – Pedro Tamaroff May 21 '13 at 13:52
In my question I don't see any regulated function. – Tony Piccolo May 21 '13 at 14:59

I think you are trying to ask for the distinction between defining the Riemann integral as a limit of supremum and infimum partitions vs just using a fixed family of partitions, in this case something akin to the dyadic intervals as Osgood does. The point is that both methods yield precisely the same result when the function $f$ is continuous on the whole interval. In fact Osgood seems to prove this fact calling it the "Convergence Theorem" on the bottom of page 113 to page 114. In other words, the type of partition and which values to sample $f$ at within the partitions are irrelevant. If you try to extend this definition to badly discontinuous functions, you'll get nonsense, whereas with the usual definition, it'll imply the supremum and infimum sums do not agree thereby not being Riemann Integrable.

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One can avoid the dyadic intervals and start from any partition using a sequence of refinements with the norm converging to zero: see pdf pag. 2 (just found). But now let's come back to the question: do you know some other reference about this approach ? – Tony Piccolo May 21 '13 at 15:30
Also, referring to pdf, without continuity you cannot show that L=U, but these limits exist and do not depend from the chosen sequence of partitions. I don't see a nonsense. – Tony Piccolo May 21 '13 at 15:54
@tony piccolo: take the pro typical dirichlet function: 1 on the rationals, 0 elsewhere. This function is not Riemann integrable because the upper and lower sums do not coincide. Yet Osgolds definition when abusively applied to this function will in fact give you a value for the integral, but the value will depend on where you sample the function – Alex R. May 21 '13 at 16:47
If the function is not continuous, the convergence theorem (p. 113) is not valid. – Tony Piccolo May 21 '13 at 17:22
Yes it's not continuous as I pointed out by saying "abuse" and "nonsense". – Alex R. May 21 '13 at 17:32

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