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.
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.