62.9. SOME MAXIMAL INEQUALITIES 2099

If X (t) is continuous, the above inequality holds without this assumption. In case p > 1,and X (t) continuous, then for each t ≤ T,(∫

Ω

|X∗ (t)|p dP)1/p

≤ pp−1

(∫Ω

X (T )p dP)1/p

(62.9.32)

Proof: The first inequality follows from Theorem 62.5.2. However, it can also beobtained a different way using stopping times.

Define the stopping time

τ ≡ inf{t > 0 : X (t)> λ}∧T.

(The infimum over an empty set will equal ∞.)This is a stopping time by 62.7.5 because itis just a continuous function of the first hitting time of an open set. Also from the definitionof X∗ in which the supremum is taken over an open interval,

[τ < t] = [X∗ (t)> λ ]

Note this also shows X∗ (t) is Ft measurable. Then it follows that X p (t) is also a sub-martingale since rp is increasing and convex. By the optional sampling theorem, the se-quence given by X (0)p ,X (τ)p ,X (T )p is a submartingale. Also [τ < T ] ∈Fτ and so∫

[τ<T ]X (τ)p dP≤

∫[τ<T ]

E (X (T )p |Fτ)dP =∫[τ<T ]

X (T )p dP

By right continuity, on [τ < T ] , X (τ)≥ λ . Therefore,

λpP([X∗ (T )> λ ]) = λ

pP([τ < T ])

≤∫[τ<T ]

X (τ)p dP≤∫[X∗(T )>λ ]

X (T )p dP

Next suppose X (t) is continuous and let {τn} be a localizing sequence,

τn ≡ inf{t : X (t)> n} .

Then by continuity, Xτn is bounded because X (τn∧ t) ≤ n, and so from what was justshown,

λpP([(Xτn)∗ (T )> λ

])≤∫[(Xτn )∗(T )>λ ]

(Xτn)(T )p dP

Then (Xτn)(T ) is increasing as τn → ∞ so the result follows from the monotone conver-gence theorem. This proves the first part.

Let Xτn be as just defined. Thus it is a bounded submartingale. To save on notation, theX in the following argument is really Xτn . This is done so that all the integrals are finite. Ifp > 1, then from the first part using the case of p = 1,

∫Ω

|X∗ (t)|p dP≤∫

Ω

|X∗ (T )|p dP =∫

∞

0pλ

p−1

≤ 1λ

∫X[X∗(T )>λ ]X(T )dP︷ ︸︸ ︷

P([X∗ (T )> λ ]) dλ