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Please correct me if I'm wrong.

In terms of Riemann integrability: If we are taking into consideration Riemann integrals on a closed interval, then any continuous function is integrable.

In terms of improper integrals: continuity does not imply integrability.

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    $\begingroup$ The answers below are correct but I want to add something, given $f$ bounded on a compact interval $I$ then $f$ is Riemann integrable on $I$ if and only if it is continuous almost everywhere. $\endgroup$
    – ℋolo
    Jun 20, 2018 at 8:27
  • $\begingroup$ @holo useful! Here's a link to a discussion elsewhere on this site, though it just has pointers to other things.math.stackexchange.com/questions/238139/… $\endgroup$ Aug 7, 2018 at 18:00

3 Answers 3

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Theorem: A continuous function $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable.

Proof:

Let $f: [a,b] \rightarrow \mathbb{R}$ be a continuous function. Any function that is continuous on a compact set—such as our $f$ on $[a,b]$—is also uniformly continuous on that set$^\dagger$. This is to say, given a $\mu > 0$, we are guaranteed a $\delta > 0$ such that $|x - y| < \delta \implies |f(x) - f(y)| < \mu$ for any $x, y \in [a,b]$. Consider a partition $\mathcal{P}$ of $[a, b]$ into $n$ equal intervals of width $\displaystyle \frac{b-a}{n}$, with $n$ large enough so that $\displaystyle \frac{b-a}{n} < \delta$. Computing the difference between the upper and lower sums: \begin{align*} U(f, \mathcal{P}) - L(f, \mathcal{P}) &= \sum_{k = 1}^{n} \left(x_k - x_{k-1} \right)\Big[\operatorname{sup}\{f(x) | x \in [x_{k-1}, x_k] \} - \operatorname{inf} \{f(x) | x \in [x_{k-1}, x_k] \} \Big] \\ & \leq \left( \frac{b-a}{n} \right) \cdot n \cdot \mu \ = \ (b-a)\mu \end{align*} Given an $\varepsilon > 0$, choose $\mu$ small enough so that $\displaystyle \mu < \frac{\varepsilon}{(b-a)}$. Then $U(f, \mathcal{P}) - L(f, \mathcal{P}) < \varepsilon$, and we conclude $f$ is Riemann integrable on $[a,b]$.


$^\dagger$ See here for further discussion.

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It is worthwhile to give another proof for Riemann integrability of functions which are continuous on a closed interval.

The proof below is taken from Calculus by Spivak and I must say it is novel enough. It does not make use of uniform continuity bur rather invokes mean value theorem for derivatives.

The central idea is to show that if $f:[a, b] \to\mathbb {R} $ is continuous on $[a, b] $ then the upper and lower Darboux integrals of $f$ on $[a, b] $ are equal ie $$\overline{\int} _{a} ^{b} f(x) \, dx=\underline{\int} _{a} ^{b} f(x) \, dx$$ Now to establish the above identity Spivak considers the upper Darboux integrals as a function of the upper limit of integration. Thus following Spivak we consider the function $$J(x) =\overline{\int} _{a}^{x} f(t) \, dt$$ and show that $J'(x) =f(x) $ for all $x\in[a, b] $. Similarly we have $j'(x) =f(x) $ for all $x\in[a, b] $ where $$j(x) =\underline{\int} _{a} ^{x} f(t) \, dt$$ The derivative of function $F=J-j$ vanishes everywhere on $[a, b] $ and $F(a) =0$ so that $F$ vanishes on whole of $[a, b] $.

The key point which needs to be established here is the relation $$J'(x) =f(x) =j'(x), \forall x\in[a, b] $$ and the proof is almost the same as that of first fundamental theorem of calculus. The upper Darboux integrals enjoy the same additive property as Riemann integrals and we have $$J(x+h) - J(x) =\overline{\int} _{x} ^{x+h} f(t) \, dt$$ Further given $\epsilon >0$ the continuity of $f$ at $x$ ensures the existence of a $\delta>0$ such that $$f(x) - \epsilon<f(t) <f(x) +\epsilon$$ whenever $t\in(x-\delta, x+\delta) $. If $0<h<\delta$ then the above inequality yields $$h(f(x) - \epsilon) \leq J(x+h) - J(x) \leq h(f(x) +\epsilon) $$ or $$\left|\frac{J(x+h) - J(x)} {h} - f(x) \right|\leq \epsilon$$ The same identity holds even when $-\delta<h<0$ and hence by definition of derivative we have $J'(x) =f(x) $. The proof for $j'(x) =f(x) $ is exactly the same (using lower Darboux integrals).

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$f(x)=1/x$ is continuous on $[1, \infty)$, but $\int_1^{\infty} f(x) dx = \infty$.

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  • $\begingroup$ So does that mean my statements are correct? $\endgroup$ Jun 20, 2018 at 7:48
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    $\begingroup$ This is not a Riemann integral, though. $\endgroup$
    – egreg
    Jun 20, 2018 at 9:15
  • $\begingroup$ Why the downvote ???? In my example the functin $f$ is continuous on $[1, \infty)$ but not integrable over $[1, \infty)$. $\endgroup$
    – Fred
    Jun 20, 2018 at 11:19
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    $\begingroup$ @Fred This is an improper Riemann integral. Usually the Riemann integral (in the narrow sense) is only defined on an interval $[a, b]$. $\endgroup$
    – ComFreek
    Jun 20, 2018 at 12:42
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    $\begingroup$ $[1, \infty)$ is not a closed interval. The questions asks about Riemann integrals on closed intervals, so the counterexample in this answer is not relevant to the question. $\endgroup$
    – pts
    Jun 20, 2018 at 13:13

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