# 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], §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.

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}