1. ## problem about almost sure and L1 convergence

Let $\displaystyle \lim_{n \to \infty}X_{n}=X$ a.s.

And let $\displaystyle Y=\sup_n|X_{n}-X|$.

• Proove that $\displaystyle Y<\infty$ a.s.
• Let $\displaystyle Q$ a new probability measure defined as it follows:
$\displaystyle \displaystyle Q(A)=\frac{1}{c} \mathbb{E}\!\left[1_{A} \frac{1}{1+Y}\right]$, where $\displaystyle \displaystyle c=\mathbb{E}\!\left[\frac{1}{1+Y}\right]$.
Proove that $\displaystyle X_{n} \rightarrow X$ (in $\displaystyle L_{1}(Q)$).

2. Originally Posted by yavanna

Let $\displaystyle \lim_{n \to \infty}X_{n}=X$ a.s.

And let $\displaystyle Y=\sup_n|X_{n}-X|$.

• Proove that $\displaystyle Y<\infty$.
• Let $\displaystyle Q$ a new probability measure defined as it follows:
$\displaystyle \displaystyle Q(A)=\frac{1}{c} \mathbb{E}\!\left[1_{A} \frac{1}{1+Y}\right]$, where $\displaystyle \displaystyle c=\mathbb{E}\!\left[\frac{1}{1+Y}\right]$.
Proove that $\displaystyle X_{n} \rightarrow X$ (in $\displaystyle L_{1}$).
It should be pointed out that the first question is just a question about real sequences (i.e., not probability) : prove that a convergent sequence is bounded, and explain why it is sufficient to conclude.

As for the second question, be very careful! It is not $\displaystyle L^1$, but $\displaystyle L^1(Q)$ that you probably mean; this is very different. The only advice then is to use the bounded convergence theorem; that should do the trick pretty neatly.

3. Thanks!

Originally Posted by Laurent
It should be pointed out that the first question is just a question about real sequences (i.e., not probability) : prove that a convergent sequence is bounded, and explain why it is sufficient to conclude.

As for the second question, be very careful! It is not $\displaystyle L^1$, but $\displaystyle L^1(Q)$ that you probably mean; this is very different. The only advice then is to use the bounded convergence theorem; that should do the trick pretty neatly.
Yes, you're right... It was $\displaystyle L_{1}(Q)$.

But the convergence to be verified in the first question is a.s.

Originally Posted by yavanna

• Proove that $\displaystyle Y<\infty$ a.s.

4. Originally Posted by yavanna
But the convergence to be verified in the first question is a.s.
Yes. The assumption means that for $\displaystyle \omega$ in an event of probability 1, the sequence $\displaystyle (X_n(\omega))_n$ of real numbers converges. However, this convergence implies that the sequence is bounded. Thus, you have the inclusion of events : $\displaystyle \{\lim_n X_n= X\}\subset\{Y<\infty\}$. If the first one has probability 1, then the same holds for the second one.