# If $f:[a,b]\to R$ is bounded with points of discontinuities enclosed by finite number of sub-intervals then $f$ is Riemann integrable on $[a,b]$.

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.

• I thought that CASE 1: if point of discontinuities are finite in number then by the theorem " any bounded function f on [a,b] with a finite number of discontinuities is Riemann integrable on [a,b]" I can show the result. But what to do if CASE 2: point of discontinuities are infinite in number. Also if we have to consider both cases, its tedious. – user1942348 Mar 4 '18 at 11:43
• How do you prove the theorem? – Lerigorilla Mar 4 '18 at 11:46
• @Lerigorilla taking finite no of subinterval enclosing the finite no of pt of discontinuities and using sufficient condition of R-integrability – user1942348 Mar 4 '18 at 11:56
• Let the intervals be $δ_1,δ_2,...,δ_n.$ Then $\exists x_1,x_2,...x_n.$ with $x_k \in δ_k \forall k\in[1,n] , k\in \mathbb{N}$ such that all of the discontinuity points lie in $(x_1-\epsilon ,x_1+\epsilon )\cup (x_2-\epsilon ,x_2+\epsilon )\cup ...\cup (x_n-\epsilon ,x_n+\epsilon ). \forall \epsilon >0$ – Lerigorilla Mar 4 '18 at 12:00
• @Lerigorilla What is $x_1,x_2,...x_n$? If these are the point of discontinuity, then number is finite. Also there may be discontinuity in $(x_1+\epsilon, x_2-\epsilon)$ etc – user1942348 Mar 4 '18 at 12:02

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.

• Would you please write elaborately. Thanks. – user1942348 Mar 25 '18 at 17:26

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).

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 (, §1.7 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.

 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.

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}