Kenneth Kuttler

300 CHAPTER 12. THE CONSTRUCTION OF MEASURES

Also suppose f ,g are nonnegative measurable functions and there exists β > 1,0 < r ≤ 1,such that for all λ > 0 and 1 > δ > 0,

µ ([ f > βλ ]∩ [g≤ rδλ ])≤ φ (δ )µ ([ f > λ ]) (12.7.17)

where limδ→0+ φ (δ ) = 0 and φ is increasing. Under these conditions, there exists a con-stant C depending only on β ,φ ,r such that∫

Ω

F ( f (ω))dµ (ω)≤C∫

Ω

F (g(ω))dµ (ω) .

Proof: Let β > 1 be as given above. First suppose f is bounded.∫Ω

F ( f )dµ =∫

Ω

βfβ

)dµ ≤Cβ

∫Ω

fβ

)dµ

=Cβ

∫∞

0µ ([ f > βλ ])dν

Now using the given inequality,

= Cβ

∫∞

0µ ([ f > βλ ]∩ [g≤ rδλ ])dν

+Cβ

∫∞

0µ ([ f > βλ ]∩ [g > rδλ ])dν

≤ Cβ φ (δ )∫

∞

0µ ([ f > λ ])dν +Cβ

∫∞

0µ ([g > rδλ ])dν

≤ Cβ φ (δ )∫

Ω

F ( f )dµ +Cβ

∫Ω

F( g

rδ

)dµ

Now choose δ small enough that Cβ φ (δ )< 12 and then subtract the first term on the right

in the above from both sides. It follows from the properties of F again that

∫Ω

F ( f )dµ ≤CβC(rδ )−1

∫Ω

F (g)dµ.

This establishes the inequality in the case where f is bounded.In general, let fn = min( f ,n) . Then for n≤ λ , the inequality

µ ([ f > βλ ]∩ [g≤ rδλ ])≤ φ (δ )µ ([ f > λ ])

holds with f replaced with fn because both sides equal 0 thanks to β > 1. If n > λ , then[ f > λ ] = [ fn > λ ] and so the inequality still holds because in this case,

µ ([ fn > βλ ]∩ [g≤ rδλ ]) ≤ µ ([ f > βλ ]∩ [g≤ rδλ ])

≤ φ (δ )µ ([ f > λ ]) = φ (δ )µ ([ fn > λ ])

Therefore, 12.7.17 is valid with f replaced with fn. Now pass to the limit as n→ ∞ anduse the monotone convergence theorem.

300 CHAPTER 12. THE CONSTRUCTION OF MEASURESAlso suppose f, g are nonnegative measurable functions and there exists B >1,0<r<1,such that for alld > 0 and1>6>0,u([f > BA og <r6A)) < 9 (6) ULF > A) (12.7.17)where lims_,o, 9 (6) = 0 and @ is increasing. Under these conditions, there exists a con-stant C depending only on B,@,r such that[FE @)du(o) <c [ F (g(@))au(o).Proof: Let B > 1 be as given above. First suppose f is bounded.[,ronan= [,F(p pss (5)=¢ [a ([f > Bal) dNow using the given inequality,— cp fu (if > BA] lg < rda})dv+¢p [uw (If > BA] [eg > rda})avCpo(5 fr es ([e > rSA]) dvCp (8 ) fru )du+Cp fF (4 duIAIANow choose 6 small enough that Cg¢@ (6) < 5 and then subtract the first term on the rightin the above from both sides. It follows from the properties of F again thatF(f < F(g5 [F fd < CBC os) + fFThis establishes the inequality in the case where f is bounded.In general, let f, = min(f,n). Then for n < A, the inequalityu([f > BAlOl[g <r64)) < 9 (6) uM (LF > A])holds with f replaced with j;, because both sides equal 0 thanks to B > 1. If n > A, then[f >A] =[f, > A] and so the inequality still holds because in this case,M([fn > BA]N|[g<réa]) < wf > BAln[g <réA))< 9(6)M(IF>A]) = 9 (5) HE (fn > Al)Therefore, 12.7.17 is valid with f replaced with f,. Now pass to the limit as n — co anduse the monotone convergence theorem. §f