If $f:[a,b]\to R$ is bounded such that its points of discontinuities can be enclosed by finite number of sub-intervals with total length arbitrarily small then show that $f$ is Riemann integrable on $[a,b]$.
I know that any bounded function $f$ on $[a,b]$ with a finite number of discontinuities is Riemann integrable on $[a,b]$. But how to show the above.
On
Denote by $D:=\{x\in[a,b]:\hspace{0.1cm}f(x)\hspace{0.1cm}\text{is discontinuous}\}$. First we show that $D$ is a set of measure zero. By the hypothesis for every $\varepsilon>0$ there exists a finite number of subintervals $\{[a_k,b_k]\}$ such that $$D\subseteq\bigcup_k[a_k,b_k]\hspace{0.2cm}\text{and}\hspace{0.2cm}\sum_kv([a_k,b_k])<\varepsilon$$ where $v([a_k,b_k])$ is the volume (the ordinary Lebesgue measure) of the subinterval $[a_k,b_k]$. Since $\varepsilon$ can be made arbitrarily small then by definition of a set of measure zero it follows that $D$ is of measure zero. This implies that $f$ is continuous almost everywhere. Additionally by assumption $f$ is bounded. Therefore $f$ is Riemann integrable (any bounded and almost everywhere continuous function on some compact interval $[a,b]$ is Riemann integrable).
On
I answer the question in two steps: in the first one I prove that the class functions $f$ in question are Riemann integrable even if not bounded. To see this let's recall the following definitions
Definition 1. The measure of an interval $[a,b]$ is its length $\mu ([a,b])=\ell ([a-b])=a-b$.
Definition 2. A set $S\subset\mathbb{R}$ is said to be a set of measure zero if, given any $\varepsilon>0$ there exists a countable (i.e finite or denumerable) family of intervals $\{[a_n,b_n]\}_{n\in\mathbb{N}}$ which covers $S$ and whose total length is less than $\varepsilon$.
According to definition 2, the set $D$ of discontinuity points of $f$ is a set of measure zero, since as already noted by Arian, $$ D\subseteq\bigcup_{0\leq n\leq N(\varepsilon)}[a_n,b_n]=D_\varepsilon $$ with $$ \mu(D_\varepsilon)=\sum_{n=0}^{N(\varepsilon)} \mu ([a_n,b_n]) = \sum_{n=0}^{N(\varepsilon)}\ell ([a_n-b_n])<\varepsilon \quad \forall \varepsilon>0 $$ where $N(\varepsilon)$ is the (finite) cardinality of the covering family. It necessary to consider it as a function $N:\mathbb{R}_+\to \mathbb{N}$ since, from the hypothesis given in the question, we only know that the discontinuity set of $f$ is covered by a finite number of intervals: this is true even when $D$ it is a denumerable set with a finite number of accumulation points, but in this case $N$ increases as $\varepsilon$ goes to $0$. Now the crucial point: for every $0<\varepsilon<b-a$, $$ [a,b]\setminus D_\varepsilon\text{ is the union of a finite number of (half open or open) intervals.} $$ This fact can be easily checked if the intervals in $\{[a_n,b_n]\}_{0\leq n\leq N(\varepsilon)}$ are pairwise disjoint or if they, on the contrary, overlap each other: to see that this fact is true even in the intermediate cases, it suffices to consider the connected components of $\bigcup_{0\leq n\leq N(\varepsilon)}[a_n,b_n]$. Therefore $f$ is Riemann integrable in every set $[a,b]\cap D_\varepsilon$ since it is continuous on each of its finite number of subintervals: passing to the limit we get $$ \int\limits_{[a,b]\setminus D_\varepsilon} f(x)\mathrm{d}x\underset{\varepsilon\to 0}\longrightarrow \int\limits_{[a,b]} f(x)\mathrm{d}x. \tag{1} \label{1} $$ Thus, under the hypothesis of the question, $f$ is Riemann integrable even without assuming its boundedness.
The second step is to note that, while $f$ is R-integrable, the limit in \eqref{1} can possibly be $\infty$: however, if $f$ is also bounded, this possibility is ruled out, since if $M\geq0$ is the supremum of $|f|$, then $$ \left\vert\:\int\limits_{[a,b]\setminus D_\varepsilon} f(x)\mathrm{d}x\right| \leq M\!\!\!\!\!\!\int\limits_{[a,b]\setminus D_\varepsilon}\!\!\!\!\mathrm{d}x < M (b-a), $$ and thus $$ \int\limits_{[a,b]} f(x)\mathrm{d}x<M(b-a) \tag{2} \label{2} $$
Notes
Lebesgue's criterion for Riemann integrability ([1], §1.7 p.20) states that if $f:[a,b]\to \mathbb{R}$ a real function of one variable, continuous outside a set $D\subset[a,b]$ $$ f\text{ is Riemann integrable }\iff \mu(D)=0 $$
The result is a consequence of the general properties of Riemann's integral and is true even without the hypothesis of boundedness of $f$: it implies that the R-integrability of a function doe not depend on the cardinality of its discontinuity set nor on other topological properties it may have.
[1] Shilov, G. E. and Gurevich, B. L. (1977), Integral, Measure, and Derivative: A Unified Approach, revised edition, Dover books on advanced mathematics, New York: Dover Publications, pp. xiv+233, ISBN 0-486-63519-8, MR 0466463, Zbl 0391.28007.
On
The following does not refer to Lebesgue measure or integration.
I shall use the following definition: An $f:\>[a,b]\to{\mathbb X}\in\{{\mathbb R}, {\mathbb C},{\mathbb R}^n\}$ is Riemann integrable over $[a,b]$ if for every $\epsilon>0$ there is a partition ${\cal P}$ of $[a,b]$ into subintervals $I_k$ $(1\leq k\leq N)$ of length $|I_k|$ such that $$D_{\cal P}(f):=\sum_{k=1}^N |\Delta f|_{I_k}\>|I_k|<\epsilon\ ,$$ whereby $|\Delta f|_I:=\sup_{x,\>y\in I}|f(x)-f(y)|$ measures the "range width" of $f$ on $I$. If $f$ is real-valued the quantity $D_{\cal P}(f)$ is just the difference between the upper and the lower Riemann sum of $f$ associated to ${\cal P}$.
Given an $f$ as in the question, and an $\epsilon>0$, there is an $M$ with $|f(x)|\leq M$ for all $x\in[a,b]$, and a partition ${\cal P}$ of $[a,b]$ such that the discontinuities of $f$ are contained in the interiors of certain "bad" intervals $I_k$ of total length $<{\epsilon\over2M}$. The union of the "good" intervals forms a compact set, on which $f$ is uniformly continuous. We can therefore refine ${\cal P}$ on these good intervals such that $$|\Delta f|_{I_k}<{\epsilon\over 2(b-a)}$$ on all resulting subintervals $I_k$ of the refined partition, again denoted by ${\cal P}$. We then have $$\eqalign{D_{\cal P}(f)&=\sum_{{\rm bad\ } I_k}|\Delta f|_{I_k} |I_k|+\sum_{{\rm good\ } I_k}|\Delta f|_{I_k} |I_k|\cr &\leq 2M \sum_{{\rm bad\ } I_k}|I_k|+{\epsilon\over2(b-a)}\sum_{{\rm good\ } I_k}|I_k|\quad\leq\epsilon\ .\cr}$$
Choose a set $A$ whose complement is a finite union on intervals with total length at most $\epsilon_1$, and such that $f$ is continuous on $A$. Then $A$ is a finite union of intervals.
On $A$, find partitions with a lower and upper sum for $\int_Af$ that are very close, say of difference at most $\epsilon_2$. (They exist by Riemann integrability of $f$ on $A$.)
Extend the partition trivially to a partition of the entire interval $[a, b]$. If $f$ is bounded by $M$, the upper and lower sum for the extended partition change with $\epsilon_1 M$ at most.
Then the upper and lower sum for the extended partition have difference at most $\epsilon_2+2M\epsilon_1$. Now take $\epsilon_1=\epsilon_2$ arbitrarily small.