Comment by verylongname | Hacker News Reader

Comment by verylongname | Hacker News Reader

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Lebesgue integrals differ in several key respects from Riemann integrals. Intuitively, Riemann integrals involve partitioning the domain of a function into disjoint intervals

(a_1,a_2), (a_2,a_3), ... (a_{n-1},a_n)

and approximately the area under the graph of the function via sums of the form

\sum_i f(x_i) (a_{i+1} - a_i)

where x_i is a point in the interval (a_i,a_{i+1}).

Lebesgue integrals turn this procedure on its head by partitioning the range of f into disjoint intervals

(a_0,a_1), (a_1, a_2), ... (a_{n-1},a_n)

and approximating the area under the graph of the function via

\sum_i m({x : a_i < f(x}) < a_{i+1}) a_i

where m(E) refers to the "measure" of the set E. That is, for each i, we multiply the "size of the set on which f is mapped to a value near a_i" by a_i and then sum over i.

The principle advantage of the Lebesgue scheme is that f can be very badly behaved and the quantities involved are still well-defined and make sense, whereas the Riemann integral only leads to reasonable approximations if f is somewhat well-behaved (more-or-less continuous). Otherwise, the value of f(x_i) (x_{i+1}-x_i) is not a reasonable approximation of the area under the graph of f "over the interval (x_{i+1}-x_i)".

There are even more general notions of integral. To my knowledge, most are based on observing that an integral is a linear functional on some space which should satisfy certain properties.

This is a nice summary, but I've always found the characterization of Riemann integrals as "partitioning the domain" and Lebesgue integrals as "partitioning the range" unsatisfying. This is mainly an artifice of the common constructions, but one can give definitions of the Riemann and Lebesgue integrals where the only difference is that, in several places, one must replace the word "finite" with the word "countable". Here is one such:

The definition of the length of an interval should be obvious. Let I be an interval, and let E be a subset of I. Define its Jordan outer measure to be the inf of the sums of the lengths of finite collections of intervals covering E. Define its Jordan inner measure to be the length of I minus the Jordan outer measure of I \setminus E. E is called Jordan measurable if its outer and inner Jordan measures are equal. A function s is Jordan simple if it is a linear combination of characteristic functions of Jordan measurable sets. Define the integral of Jordan simple functions in the obvious way. A bounded function on I is Riemann integrable if and only if it is the uniform limit of Jordan simple functions, and its Riemann integral is the limit of the integrals of the approximating simple functions.

If, in the previous paragraph, one replaces the word "finite" with the word "countable", and the names "Jordan" and "Riemann" with "Lebesgue", one recovers the Lebesgue integral.

You are just defining the class of Riemann integrable and Lebesgue integrable functions using Jordan measurability. There is nothing wrong in this, but there is nothing different in it either. It is equivalent to the standard limit of simple functions definitions.

I agree, I was just pointing out that it is possible to define both the Riemann and Lebesgue integrals by "partitioning the range". The real difference lies in the choice to allow countable rather than finite covers by intervals.

The main advantage of the Lebesgue integral isn't its generality, but the convergence theorems which give you L^p spaces. The gauge integral, despite being more general, doesn't have these properties and nobody uses it.

But the only obstruction to developing convergence theorems for Riemann integrals is that the limit of a sequence of Riemann integrable functions need not be Riemann integrable. This, of course, follows from the standard convergence theorems for Lebesgue integrals and the fact that the two notions coincide where both are defined. So it really does come down to the class of functions which are integrable.

Of course. I'm saying the important part is not that the class is bigger, the important part is that it's a Banach space.

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