列试求∑(R)ax 解:令()=a G2x∈ 则(x)为非负连续函数,当然为非负可测函数, 从而∑(R)儿ax x x x n= 1(1+x J(21=2 定理:若(x)在[ab]上 Riemann可积,则f(x)在 ab上 Lebesgue可积,且L (L)L, f(x)dx=(r)l f(x)dx例 试求 dx x x R n   n  =1 − + 1 1 2 2 (1 ) ( ) : ( ) , [ 1,1] (1 ) 2 2 =  − + f x n x x x 解 令 n dx x x R n   n  =1 − + 1 1 2 2 (1 ) 从而 ( ) dx x x L n   n  = − + = 1 [ 1,1] 2 2 (1 ) ( ) −   = + = [ 1,1] 1 2 2 (1 ) ( ) dx x x L n n ( ) 1 2 [ 1,1] = = − L dx f (x) 则 n 为非负连续函数，当然为非负可测函数， 定理：若f(x)在[a,b]上Riemann可积，则f(x)在 [a,b]上Lebesgue可积，且   = b a b a (L) f (x)dx (R) f (x)dx [ , ]