63.6. ANOTHER LIMIT FOR QUADRATIC VARIATION 2147

The inequality 63.6.24 shows that this definition is well defined. So what are the prop-erties of the integral just defined? Each

∫ t0 fndM is a continuous martingale because it is

the sum of continuous martingales. Since convergence happens in L2 (Ω) , it follows that∫ t0 f dM is also a martingale. Is it continuous? By the maximal inequality Theorem 62.9.4,

it follows that

P

([sup

t∈[0,T ]

∣∣∣∣∫ t

0fmdM−

∫ t

0fndM

∣∣∣∣> λ

])≤ 1

λ2 E

(∣∣∣∣∫ T

0( fm− fn)dM

∣∣∣∣2)

≤ 1

λ2 E(∫ T

0∥ fm− fn∥2 d [M]

)and it follows that there exists a subsequence, still called n such that for all p positive,

P

([sup

t∈[0,T ]

∣∣∣∣∫ t

0fn+pdM−

∫ t

0fndM

∣∣∣∣> 1n

])< 2−n

By the Borel Cantelli lemma, there exists a set of measure zero N such that for ω /∈ N,{∫ t0 fndM

}is a Cauchy sequence. Thus, what it converges to is continuous in t for each

ω /∈ N and for each t, it equals∫ t

0 f dM a.e. Hence we can regard∫ t

0 f dM as this continuousversion.

What is an example of such a function in GM?

Lemma 63.6.3 Let R : H→ H ′ be the Riesz map.

⟨R f ,g⟩ ≡ ( f ,g)H .

Also suppose M is a uniformly bounded continuous martingale with values in H. ThenRM ∈ GM .

Proof: I need to exhibit an approximating sequence of elementary functions as de-scribed above. Consider

Mn (t)≡mn−1

∑i=0

M (ti)X(tni ,t

ni+1]

(t)

Then clearly RMn (ti)M∗ ∈ L∞ (Ω) and so in particular it is in L2 (Ω) . Here

limn→∞

max{∣∣tn

i − tni+1∣∣ , i = 0, · · · ,mn

}= 0.

Say M∗ (ω)≤C. Furthermore, I claim that

limn→∞

E(∫ T

0∥RMn−RM∥2 d [M]

)= 0. (63.6.26)

This requires a little proof. Recall the description of [M] (t) . It was as follows. You con-sidered

Pn (t)≡ 2 ∑k≥0

((M(t ∧ τ

nk+1)−M (t ∧ τ

nk)),M (t ∧ τ

nk))

63.6. ANOTHER LIMIT FOR QUADRATIC VARIATION 2147The inequality 63.6.24 shows that this definition is well defined. So what are the prop-erties of the integral just defined? Each fo JndM is a continuous martingale because it isthe sum of continuous martingales. Since convergence happens in L* (Q), it follows thatfo fdM is also a martingale. Is it continuous? By the maximal inequality Theorem 62.9.4,it follows that2maM — ndM| >A < +e n) dM(| sue fam [na] >a} ) < be (| ummm<5 ( [iim —fi?a(m)and it follows that there exists a subsequence, still called n such that for all p positive,o( [ fuvvam - [toa > i)) <2"By the Borel Cantelli lemma, there exists a set of measure zero N such that for o ¢ N,{ Jo trdM } is a Cauchy sequence. Thus, what it converges to is continuous in ¢ for each@ ¢ N and for each, it equals {j fdM a.e. Hence we can regard fj fdM as this continuousversion.What is an example of such a function in Yy?supte [0,7]Lemma 63.6.3 Let R: H — H' be the Riesz map.(Rf,g) = (f,8)HAlso suppose M is a uniformly bounded continuous martingale with values in H. ThenRM € Gy.Proof: I need to exhibit an approximating sequence of elementary functions as de-scribed above. ConsiderThen clearly RM, (t;) M* € L® (Q) and so in particular it is in L? (Q) . Heredim max {|17 — tt |,i=0,--+ mn} =0.Say M* (q@) < C. Furthermore, I claim thatTlim E (/ \|RM, —RM|2d m)) ~0. (63.6.26)n—oo 0This requires a little proof. Recall the description of [M] (t). It was as follows. You con-sidered=2) ((M (tA th.) —M (tA t%)) M (tA t))k>0