Integrable dominating function for stopped Brownian motion [duplicate]

Possible Duplicate:
Dominated convergence problems with Wald's identity for the Brownian Motion

Let $(B_t)_{t \geq 0}$ be a Brownian motion, $T$ a stopping time with $E(T)<\infty$ and use the notation $T \wedge n$ for $\min(T,n)$.

For the proof of an identity I need to find an integrable dominating function for $B^2_{T \wedge n}$.

I tried going for $\sup_{n \in \mathrm{N}} B^2_{T \wedge n}$ and thought of using Doob's inequality for that, but I don't know if $$E(\sup_{t \geq 0} B^2_{t \wedge T \wedge n}) \leq 4 E(B^2_{T \wedge n}) \leq 4 E(B^2_{T}) < \infty$$ helps me in this regard. Any hints would be much appreciated.

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marked as duplicate by Mariano Suárez-Alvarez♦May 5 '12 at 2:00

After leaving the problem for a while, I found the rather obvious solution on reinspection.

The integrable function dominating $B^2_{T \wedge n}$ that I was looking for is $\sup_{t \geq 0} B^2_{T \wedge t}$.

We have $B^2_{T \wedge n} \leq \sup_{t \geq 0} B^2_{T \wedge t} \forall n \in \mathrm{N}$ and by Doob's inequality $E(\sup_{t \geq 0} B^2_{T \wedge t}) \leq 4 E(B^2_T) < \infty$.

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You used exactly the same text to answer another question here math.stackexchange.com/questions/136971/… ARe they the same question? –  Mariano Suárez-Alvarez May 5 '12 at 0:51
@Mariano yes the two questions are the same. –  Chris Janjigian May 5 '12 at 1:13
I'll just close this one as a dup of the other—instead of answering twice, you could have simply closed one :) –  Mariano Suárez-Alvarez May 5 '12 at 2:00
@MarianoSuárez-Alvarez Thanks, I am not very familiar with the system yet. Sorry for any inconvenience I should have caused. –  Hans-Peter Schrei May 5 '12 at 8:06
Yes, it is the same question, but with less baggage than the other one. After nobody answered the first one, I thought that its length was rather forbidding. Therefore I tried to shorten it. –  Hans-Peter Schrei May 5 '12 at 8:10