32.3. THE QUADRATIC VARIATION 865

Next suppose σn is a localizing sequence for the local martingale(submartingale) M.Then define

ηn ≡ inf{t > 0 : ∥M (t)∥> n} .

Therefore, by continuity of M, ∥M (ηn)∥ ≤ n. Now consider τn ≡ ηn ∧σn. This is anincreasing sequence of stopping times. By continuity of M, it must be the case that ηn→∞.Hence σn∧ηn→ ∞.

Finally, consider the last claim. Pick ω. Then X (τn (ω)∧δ (ω))(ω) is eventually con-stant as n→ ∞ because for all n large enough, τn (ω) > δ (ω) and so this sequence offunctions converges pointwise. That which it converges to, denoted by X (δ ) , is Fδ mea-surable because each function ω → X (τn (ω)∧δ (ω))(ω) is Fδ∧τn ⊆ Fδ measurable.■

Observation 32.3.4 Suppose M is a local martingale and τn is a localizing sequenceof stoppings times. Does Mτn (t) converge in probability to M (t)? Mτk (t) = M (t) at ω

where τk (ω) = ∞ and so [∥Mτk (t)−M (t)∥> ε] ⊆ [τk < ∞] and P([τk < ∞])→ 0 by as-sumption that τk is a localizing sequence.

One can also give a generalization of Lemma 32.2.1 to conclude a local martingalemust be constant or else they must fail to be of bounded variation.

Corollary 32.3.5 Let Ft be a normal filtration and let A(t) ,B(t) be adapted to Ft ,continuous, and increasing with A(0) = B(0) = 0 and suppose A(t)−B(t) ≡ M (t) is alocal martingale. Then M (t) = A(t)−B(t) = 0 a.e. for all t.

Proof: Let {τn} be a localizing sequence for M. For given n, consider the martingale,

Mτn (t) = Aτn (t)−Bτn (t)

Then from Lemma 32.2.1, it follows Mτn (t) = 0 for all t for all ω /∈ Nn, a set of measure0. Let N = ∪nNn. Then for ω /∈ N, M (τn (ω)∧ t)(ω) = 0. Let n→ ∞ to conclude thatM (t)(ω) = 0. Therefore, M (t)(ω) = 0 for all t. ■

Recall Example 31.3.13 on Page 841. For convenience, here is a version of what itsays.

Lemma 32.3.6 Let X (t) be continuous and adapted to a normal filtration Ft and let η

be a stopping time. Then if K is a closed set,

τ ≡ inf{t > η : X (t) ∈ K}

is also a stopping time.

Proof: First consider Y (t) = X (t ∨η)− X (η) . I claim that Y (t) is adapted to Ft .Consider U and open set and [Y (t) ∈U ] . Is it in Ft? We know it is in Ft∨η . It equals

([Y (t) ∈U ]∩ [η ≤ t])∪ ([Y (t) ∈U ]∩ [η > t])

Consider the second of these sets. It equals

([X (η)−X (η) ∈U ]∩ [η > t])