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I have been looking at this proof in Kallenberg's "Foundations of Modern Probability":

I can see that the Banach-Alaoglu theorem is being implicitly referenced, and I believe I understand the proof up to the last line:

By approximation it follows easily that $\xi_n \to \xi$ weakly in $L^1$ along $N'$.

My attempt to complete this last part of the proof goes like this:

Fix $\epsilon>0$. Since $\eta_k\to\xi$ in $L^1$, for sufficiently large $k$ we have $E|\eta_k-\xi|<\epsilon$. Also, by the uniform integrability of $\xi_n$, for sufficiently large $k$ we have $\sup_n E|\xi_n - \xi_n^k|<\epsilon$. So, choose a $k$ such that both these inequalities hold. Then, for this choice of $k$, since $\xi_n^k \to \eta_k$ in $L^1$ as $n\to\infty$ along $N'$, there is an $M$ such that $E|\xi_n^k-\eta_k|<\epsilon$ for $n \geq M$ in $N'$. Putting this together, we obtain $$E|\xi_n-\xi| \leq E|\xi_n-\xi_n^k| + E|\xi_n^k-\eta_k| + E|\eta_k-\xi| \leq 3\epsilon$$ for $n \geq M$ in $N'$. Since $\epsilon$ was arbitrary, this shows that $\xi_n\to \xi$ in $L^1$ along $N'$.

However, something seems to be wrong, since this proves too much. We should only be able to show weak convergence in $L^1$. Where is the mistake?

Brent Kerby
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  • Look at the phrase at the end of the first paragraph: "$\xi^k_n\to\eta_k$ holds weakly in $L^2$ and then also in $L^1$". I think you are meant to read that conclusion as "also weakly in $L^1$." –  Aug 13 '17 at 00:39
  • Hmm, but I was thinking weak convergence in $L^2$ implied (strong) convergence in $L^1$: for if $X_n \to X$ weakly in $L^2$, then setting $Y = \text{sgn} (X_n -X) \in L^\infty \subseteq L^2$, we have $E(X_n Y) \to E(XY)$, hence $E|X_n-X| = E((X_n - X)Y) = E(X_nY) - E(XY) \to 0$ as $n\to\infty$, hence $X_n \to X$ in $L^1$. But I know something must be wrong with this again, because here $Y\in L^\infty \cong (L^1)^*$, so this same argument would show that weak convergence in $L^1$ implies (strong) convergence in $L^1$, which can't be right. – Brent Kerby Aug 13 '17 at 01:09
  • $\mbox{sgn}(X_n-X)$ depends on $n$! –  Aug 13 '17 at 01:16
  • Oh, of course. Thanks! – Brent Kerby Aug 13 '17 at 01:17
  • This completely resolves my confusion, and I see how to finish the proof now. If you'd like to put your comment in an answer I'd be happy to accept it! – Brent Kerby Aug 13 '17 at 01:25

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Consider the phrase at the end of the first paragraph:

$\xi^k_n\to \eta^k$ holds weakly in $L^2$ and then also in $L^1$.

I think you are meant to read that conclusion as "also weakly in $L^1$."