A detail in proving that a sequence of uniformly integrable random variables converges weakly.

Question

I would like to prove the following result that is stated in a textbook, but I am not able to grasp a couple of details in the proof.

Let $\{X_n: n \in N\} $ be a sequence of uniformly integrable random variables on some probability space . There there is a subsequence which converges weakly in $L^1(P); $ i.e., there is a random variable $X $ such that $$ E[X_nZ] \to E[XZ] $$ for every random variable $Z\in L^\infty(P). $

The idea provided is to start with a sequence of random variables like $$ X_{n,k} = X_n\cdot 1_{\{|X_n|\le k\}} $$ for each $k. $ This sequence is bounded in $L^2 $ by construction and therefore there exists a subsequence which converges weakly in $L^2 $ (is there a special name for this result?) Then, and this is where I have my real issue, they asy that by the Cantor diagonalization procedure, we can extract a single subsequence $\{n_j: j \in N\} $ such that $$ X_{n_j,k} \to L_k $$ weakly for each $k. $ I fail to see how I can extract a single sequence that converges to $L_k $ starting for example from the sequence $X_{1,1}, X_{2,1}, X_{3, 1}, \ldots $ If my recollection is correct, I need to keep using subsequences of subsequences to make the Cantor diagonalization process work. But, how is a subsequence of $X_{1,2}, X_{2,2}, X_{3,2}, \ldots, $ a subsequence of the first one?

I tried everything I could think, but nothing. The rest of the proof I can see, but this Cantor "thing", I cannot.

Any insight offered greatly appreciated.

Thank you. Maurice.

PS I have barely a modicum of exposure to Functional Analysis and I wonder if this result has something to do with the Dunford-Pettis theorem.

Answer 1

Extract from $\left(X_{n,1}\right)_{n\geqslant 1}$ a subsequence $\left(X_{n_k^{(1)},1}\right)_{k\geqslant 1}$ which converges wealky to some $L_1$. Then extract from $\left(X_{n_k^{(1)},2}\right)_{k\geqslant 1}$ a subsequence $\left(X_{n_k^{(2)},2}\right)_{k\geqslant 1}$ converging weakly to some $L_2$ in $L^2$. By definition, $\{n_k^{(2)},k\geqslant 1\}\subset \{n_k^{(1)},k\geqslant 1\}$.

Now we can continue the process: suppose that we found sequences of integers $\{n_k^{(1)},k\geqslant 1\},\dots,\{n_k^{(i)},k\geqslant 1\}$ such that $$ \{n_k^{(i)},k\geqslant 1\}\subset \{n_k^{(i-1)},k\geqslant 1\}\subset\dots\subset\{n_k^{(1)},k\geqslant 1\} $$ and $\left(X_{n_k^{(u)},u}\right)_{k\geqslant 1}$ converges to some $L_u$ weakly in $L^2$. We have to find an increasing sequence of integers $(n_k^{(i+1)})_{k\geqslant 1}$ such that $\{n_k^{(i+1)},k\geqslant 1\}\subset \{n_k^{(i)},k\geqslant 1\}$ and $\left(X_{n_k^{(i+1)},i+1}\right)_{k\geqslant 1}$ converges to some $L_{i+1}$ weakly in $L^2$. To do so, extract from $\left(X_{n_k^{(i)},i+1}\right)_{k\geqslant 1}$ a weakly convergent subsequence in $L^2$.

To sum up, we found increasing sequences of integers $(n_k^{(i)})_{k\geqslant 1}$, $i\geqslant 1$ such that

• for each $i\geqslant 1$, $\{n_k^{(i+1)},k\geqslant 1\}\subset \{n_k^{(i)},k\geqslant 1\}$ and
• for each $i\geqslant 1$, $\left(X_{n_k^{(i)},i}\right)_{k\geqslant 1}$ converges to some $L_i$ weakly in $L^2$.

To conclude, let $n_k:= n_k^{(k)}$; by the first bullet, $(X_{n_k,i})$ is a subsequence of $(X_{n_k^{(i)},i})$ for all $i$.