31.2. SLICING MEASURES 1089

Then dividing by α (B(x,r)) and taking a limit as r → 0, it follows that for α a.e. x,1 = νx (Rm) , so these νx are probability measures off a set of α measure zero. Lettinggk (x) ↑ XA (x) , fk (y) ↑ XB (y) for A,B open, it follows that 31.2.10 is valid for g(x)replaced with XA (x) and f (y) replaced with XB (y).

Now let G denote the Borel sets F of Rn+m such that∫Rn+m

XF (x,y)dµ (x,y) =∫Rn

∫Rm

XF (x,y)dνx (y)dα (x)

and that all the integrals make sense. As just explained, this includes all Borel sets of theform F = A×B where A,B are open. It is clear that G is closed with respect to countabledisjoint unions and complements, while sets of the form A×B for A,B open form a π

system. Therefore, by Lemma 12.12.3, G contains the Borel sets which is the smallest σ

algebra which contains such products of open sets. It follows from the usual approximationwith simple functions that if f ≥ 0 and is Borel measurable, then∫

Rn+mf (x,y)dµ (x,y) =

∫Rn

∫Rm

f (x,y)dνx (y)dα (x)

with all the integrals making sense.This proves the theorem in the case where f is Borel measurable and nonnegative.

It just remains to extend this to the case where f is only µ measurable. However, fromregularity of µ there exist Borel measurable functions g,h,g≤ f ≤ h such that∫

Rn+mf (x,y)dµ (x,y) =

∫Rn+m

g(x,y)dµ (x,y)

=∫Rn+m

h(x,y)dµ (x,y)

It follows ∫Rn

∫Rm

g(x,y)dνx (y)dα (x) =∫Rn

∫Rm

h(x,y)dνx (y)dα (x)

and so, since for α a.e. x,y→ g(x,y) and y→ h(x,y) are νx measurable with

0 =∫Rm

(h(x,y)−g(x,y))dνx (y)

and νx is a Radon measure, hence complete, it follows for α a.e. x, y→ f (x,y) must beνx measurable because it is equal to y→ g(x,y) , νx a.e. Therefore, for α a.e. x, it makessense to write ∫

Rmf (x,y)dνx (y) .

Similar reasoning applies to the above function of x being α measurable due to α beingcomplete. It follows∫

Rn+mf (x,y)dµ (x,y) =

∫Rn+m

g(x,y)dµ (x,y)

=∫Rn

∫Rm

g(x,y)dνx (y)dα (x)

=∫Rn

∫Rm

f (x,y)dνx (y)dα (x)

with everything making sense.