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By Victor Guillemin

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34) which shows that f is integrable. We have shown that continuity is sufficient for integrability. However, continuity is clearly not necessary. What is the general condition for integrability? To state the answer, we need the notion of measure zero. 9. Suppose A ⊆ Rn . The set A is of measure zero if for every � > 0, there exists a countable covering of A by rectangles Q1 , Q2 , Q3 , . . such that � i v(Qi ) < �. 10. Let f : Q → R be a bounded function, and let A ⊆ Q be the set of points where f is not continuous.

The support of f is supp f = {x ∈ U : f (x) �= 0}. 164) For example, supp fQ = Q. 27. Let f : U → R be a continuous function. The function f is compactly supported if supp f is compact. Notation. C0k (U ) = The set of compactly supported C k functions on U . 165) Suppose that f ∈ C0k (U ). Define a new set U1 = (Rn −supp f ). Then U ∪U1 = Rn , because supp f ⊆ U . 166) 0 on U1 . The function f˜ is C k on U and C k on U1 , so f˜ is in C0k (Rn ). So, whenever we have a function f ∈ C k is compactly supported on U , we can drop the tilde and think of f as in C0k (Rn ).

We do not prove the other direction. Corollary 4. Suppose f : Q → R is R. integrable and that f ≥ 0 everywhere. If � f = 0, then f = 0 except on a set of measure zero. Q 38 Proof. Let D be the set of points where f is discontinuous. The function f is R. integrable, so D is of measure zero. If p ∈ Q − D, then f (p) = 0. To see this, suppose that f (p) = δ > 0. The function f is continuous at p, so there exists a rectangle R0 centered at p such that f ≥ nδ/2 on R0 . Choose a partition P such that R0 is a rectangle belonging to P .

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