Integration defined as a sum can be defined in different ways:
(Top) The Riemann–Darboux approach -- Divide the base of the mountain into a grid of 1 meter squares. Measure the altitude of the mountain at the center of each square. The volume on a single grid square is approximately 1 m2 × (that square's altitude), so the total volume is 1 m2 times the sum of the altitudes.
(Bottom) The Lebesgue approach -- Draw a contour map of the mountain, where adjacent contours are 1 meter of altitude apart. The volume of earth a single contour contains is approximately 1 m × (that contour's area), so the total volume is the sum of these areas times 1 m.
The difference between the Riemann and Lebesgue approaches thus: "to compute the Riemann integral of f, one partitions the domain [a, b] into subintervals", while in the Lebesgue integral, "one is in effect partitioning the range of f ."
Questions
A. Can you give examples that
an integration is well-defined via Riemann–Darboux, but not Lebesgue approaches?
an integration is not defined via Riemann–Darboux, but well-defined via Lebesgue approaches?
an integration is not defined via Riemann–Darboux nor Lebesgue approaches, but can still be defined via other approaches?
B. Are there more other novels ways to evaluate or define the integration, differed from 1. Riemann–Darboux and 2. Lebesgue approaches?
Source: https://en.wikipedia.org/wiki/Lebesgue_integration#Intuitive_interpretation
Theorem. Any function $f:[a,b]\to\mathbb R$ that is Riemann integrable is also Lesbesgue integrable.
However, for improper integrals, it is possible that something is Riemann integrable but not Lesbesgue integrable. The classic example is $$\int_1^\infty\frac{\sin x}{x}\mathop{}\!\mathrm{d}x.$$
The classic example for a function that is Lesbesgue integrable but not Riemann integrable is $$f(x)=\mathbf1(x\in\mathbb Q)=\begin{cases}1&\text{if }x\in\mathbb Q\\0&\text{if }x\not\in\mathbb Q.\end{cases}$$ In any interval, $\sup f=1$ and $\inf f=0$, so the function is not Riemann integrable. However, it is Lesbesgue integrable (say over $[0,1]$) because $f\neq0$ only on a set of measure $0$, so $\int f\mathop{}\!\mathrm{d}\mu=0$.
As for other types of integrals, there's the Riemann–Stieltjes integral, which is like a half-way point between the Riemann integral and the Lesbesgue integral. To integrate $f$ over $[a,b]$ with respect to $g$, you take a partition $P=\{a=x_0<x_1<\dots<x_n=b\}$ of $[a,b]$ as with the Riemann integral, but you define your upper sum as $$U(P,f,g)=\sum_{i=1}^n\left(\sup_{[x_{i-1},x_i]}f(x)\right)\cdot\left(g(x_i)-g(x_{i-1})\right),$$ and define the lower sum similarly. If these agree, then the Riemann-Stieltjes integral $\int_a^b f\mathop{}\!\mathrm{d}g$ is defined.
So the Riemann integral is the special case of the Riemann-Stieltjes integral with $g(x)=x$. In general, we don't require $g$ to be continuous -- usually we take $g$ to be monotone/of bounded variation though.
This is very useful in probability, as we can define the expectation of a (real) random variable $X$ (with distribution function $F$) to be $$\mathbb E[X]=\int_{-\infty}^\infty x\mathop{}\!\mathrm{d}F(x).$$ This lets us calculate the expectation of more random variables -- i.e. those without densities.
As a specific example of a function that is Riemann-Stieltjes integrable, but not Riemann-Darboux integrable, try $f(x)=\mathbf1(\mathbb Q\cap[0,1/2])$. Then $\int_0^1f\mathop{}\!\mathrm{d}g$ exists for any $g$ that is constant on $[0,1/2]$.
Finally, there exists something called an Ito integral, which generalises the Riemann-Stieltjes integral. Here, we set the $f$ and $g$ from the Riemann-Stieltjes integral to be stochastic processes. This e.g. lets us integrate over Brownian motions $B_t$ (which are not of bounded variation, so we can't take the Riemann-Stieltjes integral wrt $B_t$).