31.4. MAXIMAL INEQUALITIES AND STOPPING TIMES 847

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

λpP([X∗ (T )> λ ]) = λ

pP([τ < T ])X(τ)≥λ

≤∫[τ<T ]

X (τ)p dP =∫[τ<T ]

E (X (T )p |Fτ)dP =∫[X∗(T )>λ ]

X (T )p dP

This proves 31.2 in case X is bounded. In general case, suppose X is not just right contin-uous but also continuous.

Next let {τn} be a “localizing sequence” given by

τn ≡ inf{t : X (t)> n} .

If t < τn, then X (t) ≤ n by definition of τn. Could X (τn) > n? If so, then by continuity,X (t) > n for some t < τn so τn was not chosen correctly. Thus Xτn is bounded becauseX (τn∧ t)≤ n, and so from what was just shown,

λpP([(Xτn)∗ (T )> λ

])≤∫[(Xτn )∗(T )>λ ]

(Xτn)(T )p dP

Then (Xτn)(T ) is increasing to X (T ) and (Xτn)∗ (T ) increases to X∗ (T ) as n→ ∞ so 31.2follows from the monotone convergence theorem. This proves 31.2.

Let Xτn be as just defined. Thus it is a bounded sub-martingale. To save on notation,the X in the following argument is really Xτn . This is done so that all the integrals are finite.If p > 1, then from the first part,

∫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

≤ p∫

∞

0λ

p−1 1λ

∫Ω

X[X∗(T )>λ ]X (T )dPdλ

By Lemma 29.3.13, applied to the second half of the above and using Holder’s inequality,

∫Ω

|X∗ (T )|p dP ≤ p∫

Ω

X (T )∫ X∗(T )

0λ

p−2dλdP = p∫

Ω

X (T )X∗ (T )p−1

p−1dP

≤ pp−1

(∫Ω

X∗ (T )p dP)1/p′(∫

Ω

X (T )p dP)1/p

Now divide both sides by (∫

ΩX∗ (T )p dP)1/p′ and restore Xτn for X .(∫

Ω

Xτn∗ (T )p dP)1/p

≤ pp−1

(∫Ω

Xτn (T )p dP)1/p

Now let n→ ∞ and use the monotone convergence theorem to obtain the inequality of thetheorem 31.3. ■

Here is another sort of maximal inequality in which X (t) is not assumed nonnegative.A version of this was also presented earlier.