# Does continuity always imply integrability?

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.

• 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. – Holo Jun 20 '18 at 8:27
• @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/… – Joe Corneli Aug 7 '18 at 18:00

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}) &= \left(x_k - x_{k-1} \right) \sum_{k = 1}^{n} \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.

• so are my statements correct? – Alicia White Jun 20 '18 at 8:01
• Yes indeed. The first is correct, and so is the second assuming by "integrable" you mean "the improper integral converges to a finite value". Fred's example of $\displaystyle \int_1^\infty 1/x \ \text{d}x$ works since $\displaystyle \lim_{x \rightarrow \infty} \ln(x) = \infty$. – Kaj Hansen Jun 20 '18 at 8:04

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).

• Why the downvote? – Paramanand Singh Jun 26 '18 at 12:40
• I don't think such an answer deserves a downvote. One doesn't need uniform continuity because the typical addition and comparison properties are valid also for upper and lower integrals. – Tony Piccolo Jun 26 '18 at 17:57

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

• So does that mean my statements are correct? – Alicia White Jun 20 '18 at 7:48
• This is not a Riemann integral, though. – egreg Jun 20 '18 at 9:15
• Why the downvote ???? In my example the functin $f$ is continuous on $[1, \infty)$ but not integrable over $[1, \infty)$. – Fred Jun 20 '18 at 11:19
• @Fred This is an improper Riemann integral. Usually the Riemann integral (in the narrow sense) is only defined on an interval $[a, b]$. – ComFreek Jun 20 '18 at 12:42
• $[1, \infty)$ 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. – pts Jun 20 '18 at 13:13