10.12. GOOD LAMBDA INEQUALITY 299

Note that the function t→ µ ([ f > t]) is a decreasing function. Therefore, one can makesense of an improper Riemann Stieltjes integral

∫∞

0 µ ([ f > t])dF (t) . With more work, onecan have this equal to the corresponding Lebesgue integral above.

10.12 Good Lambda InequalityThere is a very interesting and important inequality called the good lambda inequality (Iam not sure if there is a bad lambda inequality.) which follows from the above theory ofdistribution functions. It involves the inequality

µ ([ f > βλ ]∩ [g≤ δλ ])≤ φ (δ )µ ([ f > λ ]) (10.13)

for β > 1, nonnegative functions f ,g and is supposed to hold for all small positive δ andφ (δ )→ 0 as δ→ 0. Note the left side is small when g is large and f is small. The inequalityinvolves dominating an integral involving f with one involving g as described below. Asabove, ν is the Lebesgue Stieltjes measure described above in terms of F , an increasingfunction. Is there any way to see the inequality in 10.13 might make sense? Look at theexpression on the left. If δ is small enough, you might think that the intersection of the twosets would have smaller measure than µ ([ f > λ ]).

Theorem 10.12.1 Let (Ω,F ,µ) be a finite measure space and let F be a continu-ous increasing function defined on [0,∞) such that F (0) = 0. Suppose also that for everyα > 1, there exists a constant Cα such that for all x ∈ [0,∞),F (αx)≤Cα F (x) . Also sup-pose f ,g are nonnegative measurable functions and there exists β > 1, such that for allλ > 0 and 1 > δ > 0,

µ ([ f > βλ ]∩ [g≤ δλ ])≤ φ (δ )µ ([ f > λ ]) (10.14)

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

F ( f (ω))dµ (ω)≤C∫

F (g(ω))dµ (ω) .

Proof: Let β > 1 be as given above. First suppose f is bounded. This is so there can beno question of existence of the integrals.

∫Ω

F ( f )dµ =∫

ΩF(

βfβ

)dµ ≤Cβ

∫Ω

F(

)dµ.

Let ν be the Lebesgue Stieltjes measure which comes from F , (dν = dF). From Theorem10.11.6, Cβ

∫Ω

F(

)dµ = Cβ

∫∞

0 µ ([ f/β > λ ])dν = Cβ

∫∞

0 µ ([ f > βλ ])dν . Now usingthe given inequality, ∫

F ( f )dµ =

∫∞

0µ ([ f > βλ ]∩ [g≤ δλ ])dν (λ )+Cβ

∫∞

0µ ([ f > βλ ]∩ [g > δλ ])dν (λ )

≤ Cβ φ (δ )∫

0µ ([ f > λ ])dν (λ )+Cβ

∫∞

0µ ([g > δλ ])dν (λ )

≤ Cβ φ (δ )∫

F ( f )dµ +Cβ

∫Ω

F( g

δ

)dµ

10.12. GOOD LAMBDA INEQUALITY 299Note that the function t > p ([f > t]) is a decreasing function. Therefore, one can makesense of an improper Riemann Stieltjes integral {ju ([f > t]) dF (t). With more work, onecan have this equal to the corresponding Lebesgue integral above.10.12 Good Lambda InequalityThere is a very interesting and important inequality called the good lambda inequality (Iam not sure if there is a bad lambda inequality.) which follows from the above theory ofdistribution functions. It involves the inequalityU([f > BA|O[g < 6A]) < 9 (8) mM (LF > A) (10.13)for B > 1, nonnegative functions f,g and is supposed to hold for all small positive 6 and$ (5d) + 0as 6 > 0. Note the left side is small when g is large and f is small. The inequalityinvolves dominating an integral involving f with one involving g as described below. Asabove, v is the Lebesgue Stieltjes measure described above in terms of F,, an increasingfunction. Is there any way to see the inequality in 10.13 might make sense? Look at theexpression on the left. If 6 is small enough, you might think that the intersection of the twosets would have smaller measure than p ([f > A]).Theorem 10.12.1 Lez (Q,.F,U) be a finite measure space and let F be a continu-ous increasing function defined on |0,°°) such that F (0) = 0. Suppose also that for everya > 1, there exists a constant Co, such that for all x € [0,°°), F (x) < CaF (x). Also sup-pose f,g are nonnegative measurable functions and there exists B > 1, such that for allA>Oand1>6>0,u([f > BA|O[g < 6A) < 9 (5) m([F > A) (10.14)where limg_,9, ¢ (5) =0 and @ is increasing. Under these conditions, there exists a con-stant C depending only on B,@ such that[F¢@)du(o)<c | F(g(@))au(o).Q QProof: Let B > 1 be as given above. First suppose f is bounded. This is so there can beno question of existence of the integrals. fo F (f)du = foF (B fh) du <Cp oF ¢ ) du.Let v be the Lebesgue Stieltjes measure which comes from F, (a v =dF). From Theorem10.11.6, Cp JoF (f) au = Cp Jo W(LF/B > Al)dv =Ce Jo" W (Lf > Bal) dv. Now usingthe given inequality,| F(f)du=Qcp | ul ([f > BA] N[g < SA])dv(a )+Cp [a ([f > BAIN [g > 6A]) dv (A)Cp9(3) [milf > Al)av(ay+cp | u(le > da}av(a)Cp9(3) [ FNdu+cp | F &lAIA