Can Lebesgue measure be defined without the extended reals? Even if Lebesgue measure and integration can be defined without using the extended reals, it might not be a very helpful formulation. My motivation for asking this question is that I'm curious about in which ways Lebesgue integration is an extension of Riemann integration, which doesn't necessarily use extended reals.
 A: Lebesgue integration extends Riemann integration in the sense that if $f$ is Riemann-integrable on a Jordan-Measurable set $D\subset \mathbb{R}^n$, then $f$ is also Lebesgue-integrable and the two definitions coincide. Notice that in this context, there is no use of extended reals because the set $D$ is necessarily bounded, as is the function $f$.
Extended reals appear as one of the ways in which the Lebesgue theory largely extends the class of sets that can be measured and the class of functions that can be integrated. This be better understood by looking at the close relationship between Riemann integration and the Jordan theory of measure, which can be seen in the fact that  $f\geq 0$ (defined on a closed box in $\mathbb{R}^n$) is Riemann-Integrable if and only if the region under the graph of $f$ in $\mathbb{R^{n+1}}$ is Jordan-Measurable. When f $f$ is not necessarily non-negative, the relationship with Jordan Measure is still clear because then $f$ is Riemann-integrable if and only if $f^{+}$ and $f^{-}$, which are non-negative, are Riemann-integrable. So the limitations of the Riemann Integration Theory are precisely those of the Jordan Measure Theory. 
And what are those? Well, it can't measure any bounded set, which leaves out the real line and some obviously null sets such as the natural numbers, and it can't even measure all open bounded sets. Lebesgue Measure, on the other hand can measure every single Jordan-Measurable set, every open set, every countable set, a good deal of unbounded sets (for which extended reals are essential), and then some more. Its algebraic properties are also vastly superior; unlike Lebesgue-Measurable sets, while the collection of Jordan-Measurable sets is closed under finite unions, intersections, and differences, it's not closed under countable unions or intersections, or even complements. This algebraic weakness in Jordan Measure is seen as an analytic weakness in the Riemann integral by the fact that it is badly behaved under limit operations, which is one of the strongest aspects of the Lebesgue integral (see the Convergence Theorems). None of these neat ways in which it extends the Jordan Theory would be possible without allowing some sets and integrals to take the value $\infty$.
