设/:E→R记EA={x∈E:f()>a,则 [f≥a] n=1[f> (=∩E n=1 [+∞)=(a-1,+∞)(a-n,+∞) a-n [([ 1/na-1/+1a例 [ ] 1 [ ] 1 n f a n E f a E  −  =  =  设f : E → R,记E[ f a] ={x E : f (x)  a},则 ( [ a-1/n a [ , ) ( , ) 1 1 + =  − +  = n n a a ( ) [ ] 1 1 n f a n E  −  = =  ( [ , )) 1 1  − +  = n n a ( [ ( [ [ a-1/n-1 a-1/n a-1/n+1 a