A.5. ITERATED INTEGRALS 859

is integrable because it is a finite sum of integrable functions, each function in the sumbeing integrable because the set of discontinuities has Jordan content 0. (why?) Letting(x,y) = z,

φ (z) = φ (x,y) = ∑R∈Gn

∑P∈Gm

φ R×PXR′×P′ (x,y)

= ∑R∈Gn

∑P∈Gm

φ R×PXR′ (x)XP′ (y) . (1.19)

For a function of two variables h, denote by h(·,y) the function x → h(x,y) andh(x, ·) the function y→ h(x,y). The following lemma is a preliminary version of Fubini’stheorem.

Lemma A.5.3 Let φ be a step function as described in (1.18). Then

φ (x, ·) ∈R (Rm) , (1.20)∫Rm

φ (·,y) dVy ∈R (Rn) , (1.21)

and ∫Rn

∫Rm

φ (x,y) dVy dVx =∫Rn+m

φ (z) dV. (1.22)

Proof: To verify (1.20), note that φ (x, ·) is the step function

φ (x,y) = ∑P∈Gm

φ R×PXP′ (y) .

Where x ∈ R′ and this is a finite sum of integrable functions because each has set of dis-continuities with Jordan content 0. From the description in (1.19),∫

Rmφ (x,y) dVy = ∑

R∈Gn

∑P∈Gm

φ R×PXR′ (x)v(P)

= ∑R∈Gn

(∑

P∈Gm

φ R×Pv(P)

)XR′ (x) , (1.23)

another step function. Therefore,∫Rn

∫Rm

φ (x,y) dVy dVx = ∑R∈Gn

∑P∈Gm

φ R×Pv(P)v(R)

= ∑Q∈G

φ Qv(Q) =∫Rn+m

φ (z) dV. ■

From (1.23),

MR′1

(∫Rm

φ (·,y) dVy

)≡ sup

{∑

R∈Gn

(∑

P∈Gm

φ R×Pv(P)

)XR′ (x) : x ∈ R′1

}

= ∑P∈Gm

φ R1×Pv(P) (1.24)