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I am an electrical engineer with a strong interest in math. I took a courses in real analysis and abstract algebra 25 years ago. I tried to teach my self Lebesgue integration, and developed some level of understanding. When we consider a Riemann integral we compute the area under a curve, and addition and multiplication are critical to the computation. Does it make sense to consider a Lebesgue integral over an algebraic field in which addition and multiplication are replaced with other operations that are totally different but still form a field?

Ted Ersek
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    The most relevant property is that $\mathbb{R}$ is a locally compact group; such groups have a natural measure called the Haar measure, with which you can integrate. –  Jan 06 '14 at 01:10
  • Dude i am a probablist. Your question puts me to shame... – Lost1 Jan 06 '14 at 03:20

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The relevant property is being something called a locally compact group. This takes a bit of interpretation:

Let's assume that $(G, \cdot)$ is a group; that is, $G$ is a set with the operation $\cdot$, and is a group under this operation. We can then equip $G$ with a topology such that the group operations $\cdot$ and inversion are both continuous in this topology. For example, $\mathbb{R}$ is a topological group in this sense.

Now let's assume that $G$ is locally compact and Hausdorff: Every point of $G$ has a compact neighborhood. Topologically, this means that if you zoom in enough on $G$, you find something that looks like a subset of a compact space. Again, $\mathbb{R}$ is locally compact.

Now we can construct something called the Haar measure on $G$. This is a Borel measure on $G$ that's also translation invariant. This measure is also finite on compact sets and inner- and outer-regular. So it's very nice; it's also unique up to a multiplicative constant.

Once we have a measure and a $\sigma$-algebra, it's easy to figure out how to carry out integration; it's entirely analogous to the usual construction of the Lebesgue integral: Start with simple functions and extend.

It's worth mentioning that Lebesgue measure is the Haar measure for $\mathbb{R}$ under $+$.