Some properties of the Riemann integral Here are proofs of Theorems 3.3.3-3.3.5, Corollary 3.3.6 and Theorem 3.3.7 for any Riemann integrable functions on [a; b]: Because the statements in the book are for continuous functions I added 0 to the number of the theorem or corollary to distinguish it from the corresponding one in the book. Theorem 3.3.30: If f and g are Riemann integrable on [a; b] and α; β 2 R then αf + βg is Riemann integrable on [a; b] and Z b Z b Z b (αf(x) + βg(x)) dx = α f(x) dx + β g(x) dx: (1) a a a Proof: (i) If α ≥ 0; by x2.5 #8 sup αf = α sup f [c;d] [c;d] for any subinterval [c; d]⊂[a; b]: Hence for any partition P of [a; b];UP (αf) = α UP (f): Also x2.5 #8 holds for the infinimum; for any S ⊂ R inffαx : x 2 Sg = α inf S if α ≥ 0: Hence Z b inffUP (αf)g = inffαUP (f)g = α inffUP (f)g = α f(x) dx: (2) P P P a Similarly LP (αf) = α LP (f) so Z b supfLP (αf)g = α f(x) dx: (3) P a By (2), (3) and the definition of the Riemann integral, αf is Riemann integrable on [a; b] and Z b Z b α f(x) dx = α f(x) dx: (4) a a (ii) For any S ⊂ R; sup(−f) = − inf f S S R b Hence, UP (−f) = −LP (f) so infP fUP (−f)g = − supP fLP (f)g = − a f(x) dx: Simi- R b larly, supP fLP (−f)g = − a f(x) dx so −f is Riemann integrable on [a; b] and Z b Z b −f(x) dx = − f(x) dx: (5) a a Combining (4) and (5) shows that (4) holds for any α 2 R: 1 (iii) Because f and g are Riemann integrable on [a; b]; for any > 0 we can find partitions P1 and P2 such that Z b Z b f(x) dx − ≤ LP1 (f) ≤ UP1 (f) ≤ f(x) dx + (6) a a and Z b Z b g(x) dx − ≤ LP2 (g) ≤ UP2 (g) ≤ g(x) dx + . (7) a a Also, for any interval [c; d] by x2.5 #9 sup(f + g) ≤ sup f + sup g [c;d] [c;d] [c;d] so for any partition P UP (f + g) ≤ UP (f) + UP (g) (8) and similarly LP (f) + LP (g) ≤ LP (f + g): (9) Adding (6) and (7) and using (8), (9) and Lemma 1 shows that if Q = P1 [ P2; Z b Z b f(x) dx + g(x) dx − 2 ≤ LP1 (f) + LP2 (g) a a ≤ LQf + LQg ≤ LQ(f + g) ≤ UQ(f + g) ≤ UQ(f) + UQ(g) ≤ UP1 (f) + UP2 (g) Z b Z b ≤ f(x) dx + g(x) dx + 2. a a This holds for every > 0: Hence Z b Z b supfLP (f + g)g = inffUP (f + g)g = f(x) dx + g(x) dx: P P a a Thus f + g is Riemann integrable on [a; b] and Z b Z b Z b (f(x) + g(x)) dx = f(x) dx + g(x) dx: (10) a a a The theorem follows from (4), (5) and (10). Remark: This result says that the Riemann integrable functions on [a; b] form a vector space and integration is a linear operator (transformation) from this vector space to R: 2 Theorem 3.3.40: If f and g are Riemann integrable on [a; b] and f(x) ≤ g(x) for all R b R b x 2 [a; b] then a f(x) dx ≤ a g(x) dx: Proof: Because f(x) ≤ g(x); for any partition P of [a; b];UP (f) ≤ UP (g): Hence any lower bound for fUP (f)g is a lower bound for fUP (g)g: In particular, Z b Z b f(x) dx = inffUP (f)g ≤ inffUP (g)g = g(x) dx: a P P a Theorem 3.3.50: If f is Riemann integrable on [a; b] then so is jfj and Z b Z b f(x) dx ≤ jf(x)j dx: (11) a a Proof: Let > 0 and let P be a partition of [a; b] such that UP (f) − LP (f) ≤ . Let 0 0 mi = inf[xi−1;xi] f; mi = inf[xi−1;xi] jfj;Mi = sup[xi−1;xi] f; Mi = sup[xi−1;xi] jfj: There are three cases. 0 0 Case (i): If mi ≥ 0; then Mi = Mi; mi = mi so 0 0 Mi − mi = Mi − mi: 0 0 Case (ii): If Mi < 0 then Mi = −mi; mi = −Mi so 0 0 Mi − mi = Mi − mi: 0 0 Case(iii): If Mi > 0; mi < 0 then Mi = maxfMi; −mig and mi ≥ 0 so 0 0 Mi − mi ≤ maxfMi; −mig < Mi − mi: In each case 0 0 Mi − mi ≤ Mi − mi so UP (jfj) − LP (jfj) ≤ UP (f) − LP (f) ≤ (12) and, by Lemma 3, jfj is integrable. Now (11) follows from Theorems 3.3.30 and 3.3.40 since f(x); −f(x) ≤ jf(x)j: Corollary 3.3.60: If f is Riemann integrable on [a; b] then Z b f(x) dx ≤ (b − a) sup jf(x)j: (13) a [a;b] Proof: By Theorem 3.3.50 Z b Z b f(x) dx ≤ jf(x)j dx: a a 3 0 Now apply Theorem 3.3.4 to the right side with g(x) the constant function sup[a;b] jfj: Theorem 3.3.70: If f is Riemann integrable on [a; b] and a < c < b then Z b Z c Z b f(x) dx = f(x) dx + f(x) dx: (14) a a c Proof: For any partition P of [a; b]; let Pc be P if c is a point of P and the partition obtained from P by adding the point c otherwise. Let P1 be the points in Pc which are less than or equal to c; so P1 is a partition of [a; c]; and let P2 be the points that are greater than or equal to c so P2 is a partition of [c; b]: Then LP (f) ≤ LPc (f) = LP1 (f) + LP2 (f) ≤ UP1 (f) + UP2 (f) = UPc (f) ≤ UP (f) Hence supfLP (f)g ≤ supfLP (f)g + supfLP (f)g ≤ inffUP (f)g + inffUP (f)g ≤ inffUP (f)g: 1 2 P 1 P 2 P P P1 P2 1 2 R b Since the right and left ends are equal to a f(x) dx; Z b Z c Z b f(x) dx = supfLP1 (f)g + supfLP2 (f)g = f(x) dx + f(x) dx: a P1 P2 a c 4.