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Dann sind aber (˜ gi )i∈N und (h 1 L+ (φ)-Funktionen mit beschr¨ ankten Integralfolgen ∞ ˜ i dφ < h j=1 1 = 1, 2j+1 ˜ i dφ + h g˜i dφ = fi dφ < 1 + A. Nach dem 2. Schritt sind sie φ-konvergent gegen g ∈ L1 (φ) bzw. h ∈ L1 (φ), und die Integrale konvergieren gegen gdφ bzw. hdφ. ˜ i f¨ Damit ist fi = g˜i − h ur i → ∞ fast u ¨berall konvergent gegen f := g − h ∈ L1 (φ) und die Integrale konvergieren gegen f dφ. Korollar 35 (Integrierbarkeitskriterium). Seien f : Rn → R und (Ij )j∈N eine Folge von Intervallen mit folgenden Eigenschaften (i) f ≥φ 0.

Vgl. Hausaufgaben. 48 Korollar 71. Seien φ1 , φ2 und φ wie im Satz. Dann gilt (i) F¨ ur φi -integrierbare Mengen Ai ⊂ Rni ist A1 × A2 φ-integrierbar und φ(A1 × A2 ) = φ1 (A1 )φ2 (A2 ). (ii) F¨ ur φi -messbare Mengen Ai ⊂ Rni ist A1 × A2 φ-messbar. (iii) Ist N1 eine φ1 -Nullmenge, so ist N1 × Rn2 eine φ-Nullmenge. Beweis. Vgl. Hausaufgaben. Satz 72. Sei f : Rn → R eine nicht-negative integrierbare Funktion und G := (x1 , . . , xn+1 ) 0 ≤ xn+1 ≤ f (x1 , . . , xn ) ⊂ Rn+1 die Menge unter dem Graphen von f .

Dann ist auch f ◦ h ∈ L1 (µn ) und es gilt f dµn = (f ◦ h)| det h | dµn . (24) Beispiel 82 (Ellipsenfl¨ ache). Der Isomorphismus h : (x, y) → (ax, by) mit a, b > 0 bildet den Einheitkreis D auf die Ellipse mit den Halbachsen a und b ab. Deren Fl¨ache ist deshalb gleich der Fl¨ ache π des Einheitskreises mal | det h | = ab: F = χh(D)dµ2 = (χh(D) ◦ h)abdµ2 = ab χDdµ2 = abπ. Wenn h kein Automorphismus, sondern ein Diffeomorphismus, also eine stetig differenzierbare Abbildung mit stetig differenzierbarem Inversen ist, liefert die allgemeine Philosophie der 57 Differentialrechnung sofort eine Vermutung, wie man diese Formel verallgemeinern sollte: Statt der Determinante von h sollte wohl die Determinante der Ableitung von h auftreten: f dµn = (f ◦ h)| det Dh | dµn .

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