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How accurate is the following assertion?

A function is Riemann-integrable (necessarily over an interval) if:

(1) it is defined on the interval, except perhaps at the end-points, and

(2) it is bounded on the interval, and

(3) it is continuous on the interval, except at a finite number of points.

PS. Posting this as a new question to avoid editing the previous post which has already been adequately answered as it stands.

EDIT: Revised assertion based on the answers given:

A (bounded) function is Riemann-integrable (over a closed interval) if:

it is continuous on the interval, except at a finite number of points.

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Striclty speaking, the integral must be defined at the endpoints, but the value is not of importance. The function needs to be bounded, by definition. The function could be discontinuous at a countable amount of points, and yet the function might still be integrable. On the other hand, a function which is Riemann integrable is continuous at an infinte and dense set over the domain of integration. – Pedro Tamaroff Feb 15 '13 at 20:49
Which one is the interesting sentence? – Pedro Tamaroff Feb 15 '13 at 21:22
@PeterTamaroff Thanks! Will you elaborate on the interesting last sentence and/or post your comment as an answer? – Ryan Feb 15 '13 at 21:53
Give me a while, and I will get something together. – Pedro Tamaroff Feb 15 '13 at 22:34
up vote 7 down vote accepted

Strictly speaking, for the Riemann integral of the function to exist, you need the function to be defined at the endpoints as well. This is because you need to be able to define Riemann sums, where the points at which the function is evaluated might include endpoints. On the other hand, the continuity requirement can be relaxed slightly. The actual criterion (Lebesgue's criterion for Riemann integrability) is that it should be continuous almost everywhere, i.e. except on a set of Lebesgue measure $0$.

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The continuity requirement can be relaxed to, say a countable number of removable or jump discontinuities, if you don't want to mess up with measure theory.

If improper integral is allowed, boundedness is not necessary, for example $$ \int_0^1 x^{-1/2}dx $$

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Are you not allowed to comment? – Pedro Tamaroff Feb 15 '13 at 20:50
It is very hard to find an equivalent characterization without reference to the Lebesgue measure theory. Here is a complete statement: A bounded function $f$ on a bounded interval $[a,b]$ is Riemann-integrable iff it is continuous almost everywhere, i.e. continuous every where except on a set of Lebesgue measure 0. For the definition of Lebesgue measure you can refer to – user60610 Feb 15 '13 at 22:16
To get a glimpse of the difficulty to characterize this condition without measure theory, see Cantor set: A function defined on $[0,1]$ can still be Riemann integrable even if it has discontinuities on the cantor set, which is uncountable. – user60610 Feb 15 '13 at 22:21
@Tong: No measure theory is needed to prove this characterization of the Riemann integrable functions: see for instance $\S 8.5$ of The terminology "measure zero" is used there, but this is not defined in terms of Lebesgue measure but given a direct, elementary definition (we might e.g. have called such sets "negligible", but why not use the standard term?). – Pete L. Clark Feb 16 '13 at 19:06
In fact measure zero as you've defined is equivalent to the Lebesgue Measure, as I'm sure you know :). I believe that Riemann knew about the result concerning "volume 0," i.e. if the set of discontinuities admits a finite cover with the standard property. Perhaps he also knew about Lebesgue's as well, but I would think only in the one-dimensional case. – cats Feb 17 '13 at 21:05

As per the OP's request:

LEMMA Let $f$ be Riemann integrable over $[a,b]$. If $P=\{t_0,\dots,t_n\}$ is a partition of $[a,b]$ such that $$U(f,P)-L(f,P)<\frac {b-a}m$$ with $m\in\Bbb N$; then there exists an $i$ such that $M_i-m_i<\frac 1 m$ where $M_i$ and $m_i$ are the supremum and infimum of $f$ over the subinterval $[t_{i-1},t_i]$.

PROOF IF it were the case $M_i-m_i\geq \frac 1 m $ for each $i$, then

$$\begin{align}\frac{b-a}m &=&\frac 1 m \sum (t_i-t_{i-1})\\ &=&\sum \frac 1 m (t_i-t_{i-1})\\ &\leq& \sum (M_i-m_i) (t_i-t_{i-1})\\ &=&U(f,P)-L(f,P)\end{align}$$

which contradicts our hypothesis.

THEOREM Let $f$ be Riemann integrable over $[a,b]$. Define $$\Gamma =\{x\in[a,b]:f \text{ is continuous in } x\}$$

Then $\Gamma$ is infinite, and dense over $[a,b]$.

PROOF Let $P$ be a partition of $[a,b]$ such that $U-L<1$, where $U$ and $L$ are the upper and lower Darboux sums of $f$. By the previous lemma, there is an $i$ for which $M_i-m_i<1$. If $i\neq 1$ or $i\neq n$, take $a_1=t_{i-1}$,$b_1=t_i$. Else, if $i=1$ take $a_1\in(a,t_1)$ and if $i=n$ take $b_1\in (t_{n-1},b)$. It will be the case that

$$\sup_{A'}f\geq \sup_A f$$

where $A'=[t_0,t_1]$ (resp. $[t_{n-1},t_n]$) and $A=[a_1,t_1]$ (resp. $[t_{n-1},b_1]$) and also that

$$\inf_{B'} f\leq \inf_B f$$

where $B'=[t_0,t_1]$ (resp. $[t_{n-1},t_n]$ and $B=[a_1,t_1]$ (resp. $[t_{n-1},b_1]$. In any case, $a<a_1<b_1<b$ and $$\sup_{I_1}-\inf_{I_1}<1$$ where $I_1=[a_1,b_1]$.

Since $f$ is integrable over $[a,b]$, it is integrable over $[a_1,b_1]$. The very same reasoning but with $m=2$ gives us $a_2,b_2$ with $a<a_1<a_2<b_2<b_1<b$ and with $$\sup_{I_2}f-\inf_{I_2}f<\frac 1 2$$

Continuing the process gives us a sequence of nested closed intervals $\{I_n:n\in\Bbb N\}$ such that $$\sup_{I_n}f-\inf_{I_n}f<\frac 1 n$$

By Cantor's theorem, $\bigcap_{n\in\Bbb N} I_n\neq \varnothing$. Let then $\mu \in \bigcap I_n$. Since the sequences of $b_i$ and $a_i$ are stricticly monotonic, it follows that $\mu$ is none of the extrema. Let $\epsilon >0$ be given, and choose $n\in \Bbb N$ such that $\frac 1 n < \epsilon$. We have that $\mu \in (a_n,b_n)$. Let $\delta >0$ such that if $|\mu-x|<\delta$, $x\in I_n$. Since $\inf f \leq f \leq \sup f$, for each $x\in I_n$, it follows that if $|x-\mu|<\delta$ then $|f(x)-f(\mu)|<\frac 1 n<\epsilon$, and thus $f$ is continuous at $x=\mu$. Since $f$ is integrable over $[a,b]$ it also is integrable over $[a,\mu]$ and $[\mu,b]$. Set $\mu=\mu_1$. By the previous reasoning, we obtain two new different points of continuity, $\mu_2$ and $\mu_3$. Having repeated this $j$ times we would obtain $2^{j+1}-1$ continuity points. Thus $f$ cannot be continuous over a finite number of points $M$, for it would mean there exists $M$ such that $2^{j+1}-1<M$ for each $j\in \Bbb N$. It follows $\Gamma$ is infinite.

Let $x\in[a,b]$ and consider the sequence

$$\sigma_0=[a,x]$$ $$\sigma_n=\left[x-\frac{x+a}{2^n},x\right]$$

The previous argument means there exists for each $n$ a $\gamma_n\in\sigma_n$ with $\gamma_n\in\Gamma$. But then given $\epsilon>0$ we can take obtain an $N$ large enough such that $\frac{|x+a|}{2^n}<\epsilon$ whenever $n\geq N$ and thus $|x-\gamma_n|\leq \frac{|x+a|}{2^n}<\epsilon$ for $n\geq N$, that is, $\gamma_n\to x$. Since $x$ was arbitrarily chosen, $\Gamma$ is dense over $[a,b]$. This completes the proof.

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