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That sounds correct. Because for any set of measure 0 in n dimensions, there must be some set of n-1 dimensional slices that all intersect the measure 0 set in sets of measure 0 (in n-1 dimensions), so you can integrate along the slices, and then integrate over the slices. Repeating this process gives you eventually 1-parameter curves that you can integrate over because they're discontinuous only on a set of measure 0 in 1 dimension.
Of course, considering how screwed up my analysis course was, I'm sure that I'm missing some pivotal condition that only matters in really esoteric cases.

A bounded function f:R^n→R is Riemann-integrable over a cell iff it is continuous almost everywhere, i.e., the set of discontinuities has Lebesgue measure zero. There is nothing about the Riemann integral which inherently depends on the concept of interval rather than some partitionining scheme. However, a measurable function f: X→R may be discontinuous everywhere at still be Lebesgue-integrable, i.e., the preimage of every open set is measurable in X. An example of this is χ_A(x) = {1, x in A; 0, otherwise}, for, say, A = Q (rationals) over E = [0,1], which has Int_E[ χ_Q dμ ] = 1 if μ is the ordinary Lebesgue measure. Note that χ_Q is discontinuous everywhere in the reals, so that the set of its discontinuities has positive measure.

I'm not sure what you're saying here, since the Riemann integral is defined in terms of partitions in the first place. If you're integrating over an interval (cell) I, let Γ = {I_k} be a finite partition of I into nonoverlapping subintervals (cells), let |Γ| = max {μ(Ι_k)}, and pick ξ_k in I_k. Then the Riemann integral A = RInt_I[ f (x) dx ] is defined as A = lim_{|Γ|→0}[ Sum_k[ f(ξ_k) μ(I_k) ] ], if it exists. Additionally, let U_Γ = Sum_k[ [sup_{x in I_k} f(x)] μ(Ι_k) ], L_Γ = Sum_k[ [inf _{x in I_k} f(x)] μ(I_k) ]; the Darboux integral is defined as A = inf U_Γ = sup L_Γ, should the latter equality hold. The function f is Riemann-integrable iff it is Darboux-integrable, and the values coincide.


A measure on X is a function μ:M→R', where M⊂P(X) is a collection of measurable sets that includes φ and X, R' is the set of nonnegative reals and +∞, μ(φ) = 0, φ(X) > 0, and furthermore μ is σ-additive, i.e., the measure union of countably many disjoint sets is the sum of their respective measures. The "Lebesgue measure" is the standard way of measuring content (volume) in Euclidean space, but measures are rather general. If μ(X) = 1, then it is a probability measure, and the probability of some event E∈M has probability μ(E). For example, if X is the set of outcomes of rolling a fair six-sided die, then μ({1,4}) = 1/3.

Note that not all subsets of X are guaranteed to be measurable; in fact, the axiom of choice guarantees that there are subsets of R^n that are not measurable. An interesting illustration of this is the Banach-Tarski theorem, a commonly stated corollary of which is that a ball in R^3 can be partitioned into finitely many pieces are reassembled through rigid motions only into two balls, each of the same volume as the original. This is not a contradiction because the pieces have no volume (measure) at all--not zero volume, but no volume.

What abstract integration allows one to do is integrate with regard to arbitrary measure rather than the standard concept of volume in R^n. Now, X can be an arbitrary measure space, possibly with no resemblance to Euclidean space at all. As a simple situation in which one might not expect integrals to be found, let X again be the space of six-sided die rolls with some measure μ (not necessarily a fair die), and f(x) is some value attached to die roll x (not necessarily 1-6, e.g., cash payout or penalty, etc.), then the expected value of the roll is Int_X[ f dμ ], which is really just a summation of six terms in disguise. Still, this sort of formalism allows one to erase the distinction between discrete and continuous probabilities. Much more complicated applications exist, of course, but I'd rather not go into them at this time.

I believe baby Rudin covers the Lebesgue integral at the very end; papa Rudin starts with it in the very first chapter. Additional note: there are systems of probability in which probability measures are only required to be finitely additive rather than σ-additive, but they invalidate standard and useful results like the continuity theorems.