63.7. DOOB MEYER DECOMPOSITION 2153

Lemma 63.7.5 Let f be right continuous. Then f is Borel measurable. Also, if the limitfrom the left exists, then f− (x) ≡ f (x)− ≡ limy→x− f (y) is also Borel measurable. If Ais an increasing right continuous function and f is right continuous and f−, the left limitfunction exists, then if f is bounded, on [a,b], and if{

xp0 , · · · ,x

pnp

}∞

p=1

is a sequence of partitions of [a,b] such that

limp→∞

max{∣∣xp

k − xpk−1

∣∣ : k = 1,2, · · · ,np}= 0 (63.7.28)

then ∫(a,b]

f−dA = limp→∞

np

∑k=1

f(xp

k−1

)(A(xp

k

)−A

(xp

k−1

))(63.7.29)

More generally, letD≡ ∪∞

p=1

{xp

0 , · · · ,xpnp

}∞

p=1

andf− (t) = lim

s→t−,s∈Df (s) .

Then 63.7.29 holds.

Proof: For x ∈ f−1 ((a,∞)) , denote by Ix the union of all intervals containing x suchthat f (y) is larger than a for all y in the interval. Since f is right continuous, each Ix haspositive length. Now if Ix and Iy are two of these intervals, then either they must have emptyintersection or they are the same interval. Thus f−1 ((a,∞)) is of the form ∪x∈ f−1((a,∞))Ixand there can only be countably many distinct intervals because each has positive length andR is separable. Hence f−1 ((a,∞)) equals the countable union of intervals and is therefore,Borel measurable. Now

f− (x) = limn→∞

f (x− rn)≡ limn→∞

frn (x)

where rn is a decreasing sequence converging to 0. Now each frn is Borel measurable bythe first part of the proof because it is right continuous and so it follows the same is true off−.

Finally consider the claim about the integral. Since A is right continuous, a simpleargument involving the dominated convergence theorem and approximating (c,d] with apiecewise linear continuous function nonzero only on (c,d +h) which approximates X(c,d]will show that for µ the measure of Definition 63.7.4

µ ((c,d]) = A(d)−A(c) .

Therefore, the sum in 63.7.29 is of the form

np

∑k=1

f(xp

k−1

)µ ((xk−1,xk]) =

∫(a,b]

np

∑k=1

f(xp

k−1

)X(xk−1,xk]dµ