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added 234 characters in body

edited Mar 12, 2010 at 4:25

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I assume you want $f$ to be defined everywhere on $D$. In that case, it's pretty clear for an "elementary" function like $\tan x$ that $D$ needs to have holes in it, and it's totally unclear what an integral "across" these holes should mean (from a real-variable perspective). You also probably don't want to consider functions whose integrals don't exist because they diverge, since you asked about discontinuities and unboundedness is a different reason for the integral not existing. That means you should really only consider the case where $D$ is a closed interval and $f$ is bounded.

In that case, there is a theorem due to Lebesgue which states that a function on a closed interval $[a, b]$ is Riemann integrable if and only if it is bounded and also continuous almost everywhere. This is true of essentially every reasonable "elementary" function I can think of; if you can write down an "elementary" function which is discontinuous on a set of positive measure then your definition of elementary is in trouble!

Edit: The functions you listed in your comment all have the property that they are continuous on intervals where they are defined, so they'll all have the above property and so will sums, products, and compositions thereof.

I assume you want $f$ to be defined everywhere on $D$. In that case, it's pretty clear for an "elementary" function like $\tan x$ that $D$ needs to have holes in it, and it's totally unclear what an integral "across" these holes should mean (from a real-variable perspective). You also probably don't want to consider functions whose integrals don't exist because they diverge, since you asked about discontinuities and unboundedness is a different reason for the integral not existing. That means you should really only consider the case where $D$ is a closed interval and $f$ is bounded.

In that case, there is a theorem due to Lebesgue which states that a function on a closed interval $[a, b]$ is Riemann integrable if and only if it is bounded and also continuous almost everywhere. This is true of essentially every reasonable "elementary" function I can think of; if you can write down an "elementary" function which is discontinuous on a set of positive measure then your definition of elementary is in trouble!

I assume you want $f$ to be defined everywhere on $D$. In that case, it's pretty clear for an "elementary" function like $\tan x$ that $D$ needs to have holes in it, and it's totally unclear what an integral "across" these holes should mean (from a real-variable perspective). You also probably don't want to consider functions whose integrals don't exist because they diverge, since you asked about discontinuities and unboundedness is a different reason for the integral not existing. That means you should really only consider the case where $D$ is a closed interval and $f$ is bounded.

In that case, there is a theorem due to Lebesgue which states that a function on a closed interval $[a, b]$ is Riemann integrable if and only if it is bounded and also continuous almost everywhere. This is true of essentially every reasonable "elementary" function I can think of; if you can write down an "elementary" function which is discontinuous on a set of positive measure then your definition of elementary is in trouble!

Edit: The functions you listed in your comment all have the property that they are continuous on intervals where they are defined, so they'll all have the above property and so will sums, products, and compositions thereof.

answered Mar 12, 2010 at 4:19

124.2k
42
468
765

I assume you want $f$ to be defined everywhere on $D$. In that case, it's pretty clear for an "elementary" function like $\tan x$ that $D$ needs to have holes in it, and it's totally unclear what an integral "across" these holes should mean (from a real-variable perspective). You also probably don't want to consider functions whose integrals don't exist because they diverge, since you asked about discontinuities and unboundedness is a different reason for the integral not existing. That means you should really only consider the case where $D$ is a closed interval and $f$ is bounded.

In that case, there is a theorem due to Lebesgue which states that a function on a closed interval $[a, b]$ is Riemann integrable if and only if it is bounded and also continuous almost everywhere. This is true of essentially every reasonable "elementary" function I can think of; if you can write down an "elementary" function which is discontinuous on a set of positive measure then your definition of elementary is in trouble!