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I'm trying to understand Riemann integration. So let me start out by writing down the definition of Riemann integration.

A bounded function $f$ is Riemann integrable $\Leftrightarrow$ $L(f)=U(f)$ where $L(f)= \inf\{L(P,f): P \text{ is a partition of } [a,b] \}, U(f)= \sup\{U(P,f): P \text{ is a partition of } [a,b]\}$ and $U(P,f)$ and $L(P,f)$ are upper and lower Riemann sums respectively.

This is a little more general than what I have been taught in my high school where we assumed the function $f$ to be simply continuous instead of just bounded on $[a,b]$ and the definite integral would thus be limit of (any) Riemann sum.

I'm aware that every continous function on a closed interval is bounded but the converse is not true.

My question is: What really has changed when we go from "$f$ is continuous" to "$f$ is just bounded"?

My understanding: What my novice eyes can see is that, the bounded function definition allows $f$ to be discontinuous. But the real question is how often can a function be discontinuous?

If $f$ is discontinuous at finite number of points $c_1<c_2< \ldots < c_{n-1}$ in $ [a,b]$ then $f$ is continuous everywhere else, hence I can deal with these discontinuities by simply breaking the function into continous parts like so, (calling $a=c_0$ and $b= c_n$) $$I=\displaystyle\int_{a}^b f = \sum\limits_{i=1}^{n} \int_{c_{i-1}}^{c_i} f$$

If discontinuities are countably infinite i.e. $c_1<c_2<\ldots $ then I simply take the limit $\lim\limits_{n\to ∞}I$. If discontinuities are uncountable then well... I'm not sure if it's even Riemann integrable.

So in other words, my question is: Can I not simply break down Riemann integral of a bounded function to a sum of Riemann integrals of continous functions, everytime? Which would mean that using "bounded" function is really not necessary as every Riemann integral of a bounded function can be written as sum of Riemann integrals of continous functions so why not just study integrals on continuous functions.

P.S.: I'm a beginner, I don't understand measure theory or Lebesgue-something. I request you to use layman terms. Sorry if this might be a dumb question.

William
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    Well, the precise characterisation of Riemann integrable functions in terms of continuity does involve the Lebesgue measure, see for example https://math.stackexchange.com/q/238139 and the links therein. – Martin R Jun 15 '21 at 11:01
  • And yes, a Riemann integrable function can have uncountably many discontinuities: https://math.stackexchange.com/q/305200, https://math.stackexchange.com/q/1039740. – Martin R Jun 15 '21 at 11:06
  • @MartinR yes, I understand. I already looked around a lot (a lot, a really lot) and I couldn't find a simple answer. People talked about measures or something, hence this question. For instance, what does "the set of discontinuities is of zero lebesgue measure" mean in simple language? – William Jun 15 '21 at 11:09
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    Unless I am mistaken, a set has Lebesgue measure zero iff it can be covered with a countable union of intervals of arbitrary small total length. – Martin R Jun 15 '21 at 11:13
  • So how would you break “the Riemann integral of a bounded function into a sum of Riemann integrals of continous functions” if there are uncountably many discontinuities? – Martin R Jun 15 '21 at 11:22
  • @MartinR Ah, I did not know $f$ can have uncountably many discontinuities and still be Riemann integrable. Forgive me for my ignorance but let me get this straight, bounded $f$ with countable discontinuities is always Riemann integrable but with uncountable discontinuities may or may not be Riemann integrable. Am I correct in saying that? – William Jun 15 '21 at 11:30
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    Yes, because a countable set has measure zero, and an uncountable set may or may not have measure zero. – Martin R Jun 15 '21 at 11:31
  • @MartinR Secondly, using that definition of Lebesgue measure zero that you give (which I think I understand),would you be perhaps kind enough to show me an example with Thomae's function that Mr Santos has mentioned below using that definition? – William Jun 15 '21 at 11:33
  • Mr Santos has written an answer, so you can ask him for clarification. My impulse is to close the question as a duplicate of https://math.stackexchange.com/q/252470, where it is demonstrated that a Riemann integrable function can have uncountably many discontinuities. – Martin R Jun 15 '21 at 11:36

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The short answer is: no you cannot. Consider Thomae's function $f$, for instance, restricted to $[0,1]$. This function is bounded ($f(\Bbb R)\subset[0,1]$), and it is Riemann-integrable. Also, it is discontinuous at every rational number. So, in this case, you cannot break $f$ in the way that you have described.

José Carlos Santos
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  • I wonder if the question should be closed as a duplicate of, e.g., https://math.stackexchange.com/q/305200/42969. – Martin R Jun 15 '21 at 11:08
  • @MartinR I don't think so, but it's not clear-cut for me. – José Carlos Santos Jun 15 '21 at 11:10
  • Your example has countably many discontinuities. Here is an example with uncountably many discontinuities: https://math.stackexchange.com/q/252470. (Still thinking this is a duplicate.) – Martin R Jun 15 '21 at 11:38
  • I suppose I'll mark this as the best answer since it does kind of answer my question. Thanks to @MartinR's links, I've come to accept there is no easy way to say what kind of functions can be Riemann integrable. I'm still trying to wrap my head around how on earth is that function even Riemann integrable. Thanks for the help! – William Jun 16 '21 at 15:35