As an example, take $$\int_{-1}^1{1\over x}dx$$. What would that evaluate to?
Now, what about, say, $$\int_0^0{1\over x}dx$$? Is it 0, undefined, or something else?
Most of that article is passing right over my head. All I really get is that it's an alternative definition of integration, and some functions are Lebesgue integrable but not Riemann integrable, and similarly some functions are Riemann integrable but not Lebesgue integrable.
Yeah, it's complicated, I warned you. Maybe you can find a simpler explanation somewhere. But that's how you integrate functions with weird discontinuities. The canonical example is
f(x)=0, if x is rational; 1, if x is irrational.
That function is nowhere continuous, but its Lebesgue integral between 0 and n is n, because almost all its values are 1.