A less known definition of the definite integral of a continuous function 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) ?
 A: 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.
SKETCH OF PROOF
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 $\sup A=b$.
A: 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.
