Types of integration

Question

So I'm studying integration as part of a real analysis module, and I've come across the regulated integral defined on regulated functions and the Riemann integral which seems to be defined for any function $f:[a,b] \rightarrow \mathbb{R}$.

We've proved that if $f \in R[a,b]$ (regulated function on $[a,b]$) then it is Riemann integratable and the two integrals are equal.

I've also heard of Lebusgue integration and I'm just wondering if all these 'types' of integrals can be ordered in a sense that I've mentioned 3 'types' of integration above, so which of these are necessary given another and sufficient for another if that makes sense? Or am I going about this all wrong? And how many different 'types' of integration are there?

If anyone has a good understanding of this topic and can understand what I'm trying to ask then it would be good if they could share some knowledge, thanks.

Riemann integrals is not defined for any function. Famous example is $f(x)=1$ if $x$ is irrational, and $f(x)=0$ otherwise. — J1U, Feb 27 '19 at 11:06
ah yes, okay I should have realised that but I find it hard to come up with examples myself but yeah I can see why that function isn't Riemann integrable — Displayname, Feb 27 '19 at 11:07
A Riemann integrable function defined in a compact set is always Lebesgue integrable, and a Lebesgue integrable function is Riemann integrable if it is only discontinuous in a measure-zero set. — J1U, Feb 27 '19 at 11:11
Besides the ones that you have mentioned, the wikipedia article on integration does a good (but not exhaustive) job at listing some of the most used ones. — Easymode44, Feb 27 '19 at 11:12
For some important purposes the Lebesgue integral has better properties in relation to taking limits than the Riemann integral, so while they agree on most "standard" functions - where one exists so does the other and they are equal - the main point is that limits can be taken in a variety of useful situations. It is a bit like moving from the rationals to the reals for analysis (and also rather unlike) - most real numbers never appear (human beings will only ever name a finite number of reals), but having the reals there means we can guarantee that the limits we need exist. — Mark Bennet, Feb 27 '19 at 11:38
See Brian S. Thomson's answers to FTC - is the differentiability of the antiderivative a necessary hypothesis AND Wanted: A simple and didactically optimized integration theory, and the various answer to How much do we really care about Riemann integration compared to Lebesgue integration. — Dave L. Renfro, Feb 27 '19 at 16:13

score 1 · Answer 1 · answered Feb 27 '19 at 11:11

To answer your first question all integrals over a function are equal. The integral itself is a property .Riemann integral is defined for nice functions that are continuous over an interval [a,b]. Now nice functions are the class C1 that consists of all differentiable functions whose derivative is continuous; such functions are called continuously differentiable and they verify the fundamental theorem of calculus, the idea of Lebesgue integration was to verify this property for functions that are not continuous on closed intervals such as the Peano function. Riemann integral is when you measure the area under the curve by dividing your x-axis into parts and approximation the area with a lot of rectangles, to proceed with this calculation you need to measure the length of the subdivision but this doesn't work for a large class of functions hence the idea of measure theory . Lebesgue integral is a generalization that builds from measure theory so you can verify the fundamental theorem of calculus for functions that are not continuous on a closed interval.

P.S: I am an undergraduate and have a Measure theory and Lebesgue integration course this semester, hopefully someone can verify what I said and give a better answer.

It does not have to be continuously differentiable, or even continuous, for a function to be Riemann Integrable. — J1U, Feb 27 '19 at 11:16
Hey, thanks for your answer, but the function $f: [0,1] \rightarrow \mathbb{R}$ defined by $f(x) = 1$ for $x=2^{-n} n=1,2,3....$ and $f(x) = 0$ otherwise is Riemann integrable and is obviously not c.t.s differentiable — Displayname, Feb 27 '19 at 11:16
Yes, the Riemann-Integrable property doesn't require continuity but I was trying to explain why the need of Lebesgue's integral. — halius, Feb 27 '19 at 11:19

Types of integration

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