2504 CHAPTER 74. A MORE ATTRACTIVE VERSION

Proof: Consider the formula in Lemma 74.3.1.

⟨BX (t) ,X (t)⟩= ⟨BX (s) ,X (s)⟩

+2∫ t

s⟨Y (u) ,X (t)⟩du+ ⟨B(M (t)−M (s)) ,M (t)−M (s)⟩

−⟨B(X (t)−X (s)− (M (t)−M (s))) ,X (t)−X (s)− (M (t)−M (s))⟩+2⟨BX (s) ,M (t)−M (s)⟩ (74.3.9)

Now let t j denote a point of Pk from Lemma 74.0.2. Then for t j > 0,X (t j) is just the valueof X at t j but when t = 0, the definition of X (0) in this step function is X (0)≡ 0. Thus

m−1

∑j=1

⟨BX(t j+1

),X(t j+1

)⟩−⟨BX (t j) ,X (t j)

⟩+⟨BX (t1) ,X (t1)⟩−⟨BX0,X0⟩

= ⟨BX (tm) ,X (tm)⟩−⟨BX0,X0⟩

Using the formula in Lemma 74.3.1, for t = tm this yields

⟨BX (tm) ,X (tm)⟩−⟨BX0,X0⟩= 2m−1

∑j=1

∫ t j+1

t j

⟨Y (u) ,X rk (u)⟩du+

+2m−1

∑j=1

⟨BX (t j) ,M

(t j+1

)−M (t j)

⟩+

m−1

∑j=1

⟨B(M(t j+1

)−M (t j)

),M(t j+1

)−M (t j)

⟩

−m−1

∑j=1

⟨B(X(t j+1

)−X (t j)−

(M(t j+1

)−M (t j)

)),

X(t j+1

)−X (t j)−

(M(t j+1

)−M (t j)

)⟩+2∫ t1

0⟨Y (u) ,X (t1)⟩du+2⟨BX0,M (t1)⟩+ ⟨BM (t1) ,M (t1)⟩

−⟨B(X (t1)−X0−M (t1)) ,X (t1)−X0−M (t1)⟩ (74.3.10)

First consider

2∫ t1

0⟨Y (u) ,X (t1)⟩du+2⟨BX0,M (t1)⟩+ ⟨BM (t1) ,M (t1)⟩ .

Each term of the above converges to 0 for a.e. ω as k→ ∞ and in L1 (Ω). This followsright away for the second two terms from the assumptions on M given in the situation.