I have learned that every continuous, or piecewise continuous function can be Riemann integrated.

But then, are there uncontinuous functions that are Riemann integrable? And if there is, can I still understand the value of a definite Riemann integral as 'an area under a curve'? I'm not sure I can imagine a totally uncontinuous function having 'area under its function' in my head

Thanks in advance

  • $\begingroup$ no I don't think "area" would be a meaningful term in the case you ask $\endgroup$ – Guy Mar 15 '14 at 6:58
  • 1
    $\begingroup$ A function on a compact interval $[a, b]$ is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). See: en.wikipedia.org/wiki/Riemann_integrable#Integrability $\endgroup$ – Cm7F7Bb Mar 15 '14 at 7:11
  • $\begingroup$ Check Lebesgue criteria for Riemann integrability. $\endgroup$ – Mhenni Benghorbal Mar 15 '14 at 13:25

A function $f$ is Riemann integrable on the interval $[a, b]$ if the following condition holds: For any partition $P = \{ x_{0}, ..., x_{n} \}$ of $[a, b]$, we have an $\epsilon > 0$ such that $U(P, f) - L(P, f) < \epsilon$.

Here, $U(P, f) = \sum_{i=1}^{n} M_{i} \Delta x_{i}$, for $M_{i} = sup \{ f(x) : x \in [x_{i-1}, x_{i}]$. The $sup$ is the supremum, or least upper bound.

Similarly, $L(P, f) = \sum_{i=1}^{n} M_{i} \Delta x_{i}$ for $m_{i} = inf \{ f(x) : x \in [x_{i-1}, x_{i}] \}$. The $inf$ is the infimum, or greatest lower bound.

Conceptually, we are just taking Riemann sums. $U(P, f)$ is a Riemann sum where, for each given interval, we take the largest value. Similarly, $L(P, f)$ is a Riemann sum where we take the smallest value on each interval. So essentially, if we can control how much these two Riemann sums differ, we can integrate $f$.

As for conceptualizing this, I'd think about Riemann integrals in this way: you add up the rate and you get a change. That's really what Calculus is about.

  • $\begingroup$ Riemann integrable functions must be bounded on the interval of integration. $\endgroup$ – Mustafa Said Mar 15 '14 at 10:16
  1. continuous function on closed interval

  2. monotonic and bounded function on closed interval.

  3. bounded functions that satisfy 1. or 2. on every sub-interval that constitutes [a,b].



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