# Types of integration

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 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 at 11:07
• 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 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 at 11:38

• 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 at 11:16