Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is the only requirement for a function to be integrable just for it to be continuous on an entire specified interval like $[a,b]$?

The function $f(x)$ that I'm talking about is in the integrand and has the mapping $f:\mathbb{R} \rightarrow \mathbb{R}$

$$\int_a^b \! f(x) \, dx$$

What about for complex functions?

share|improve this question

4 Answers 4

up vote 15 down vote accepted

Integrability is a pretty weak condition with respect to continuity/differentiability. For instance, consider the following function $f : [0,1] \to \mathbb{R}$: $$ f ( x ) = \begin{cases} \frac{1}{q}, &\text{if }x > 0\text{ is rational and }x = \frac{p}{q}\text{ in lowest terms} \\ 0, &\text{otherwise.} \end{cases}$$ You can show that this function is continuous only at irrationals (and is right-continuous at $x=0$). It follows that $f$ is not continuous on any nondegenerate interval $[a,b]$. However the function is (Riemann) integrable, with $\int_{0}^1 f(x) dx = 0$.

share|improve this answer

A function does not even have to be continuous to be integrable. Consider the step function $f(x) = \begin{cases} 0 & x \le 0\\ 1 & x > 0 \end{cases}$. It is not continuous, but obviously integrable for every interval $[a,b]$. The same holds for complex functions.

Note that many theorems about integration will, in fact, require stronger conditions on $f$.

share|improve this answer

If you allow Lebesgue integration, then the condition is even weaker. For example, consider the function $f(x) = 0$ if $x$ is rational, $f(x)=1$ is $x$ is rational. This is continuous nowhere but integrable.

share|improve this answer

It is a theorem that a function is Riemann integrable if and only if it is bounded and its set of discontinuities has Lebesgue measure zero.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.