At the beginning of section 6.2 in "Probability Theory: A comprehensive course" Klenke states: "From the preceding section, we can conclude that convergence in measure plus existence of L1 limit points implies L1-convergence. Hence convergence in measure plus relative sequential compactness in L1 yields convergence in L1. In this section, we study a criterion for relative sequential compactness in L1, the so-called uniform integrability".
I have gone through the preceding sections and I cannot find what Klenke states. What are the theorems or facts that would lead me to Klenke's conclusion?
L1 convergence is defined as $\int |f_n-f|d\mu \xrightarrow{} 0$ as n grows to infinity. Convergence in measure is defined as $\mu(\{d(f_n,f)>\epsilon\}\cap A)\xrightarrow{} 0$ as $n\xrightarrow{} \infty$ for all $\epsilon>0$ and all $A\in\mathcal{F}$ with $\mu(A)<\infty$
I already know from Athreya about Scheffe's Theorem which requires convergence almost everywhere, convergence in integral and that the limit be integrable, for L1 convergence to result. It seems that Klenke's statement is more general and less restrictive. I suspect that fast convergence in measure may be involved.
The result, as stated, is fasle. Suppose $f_n \to f$ measure but not in $L^{1}$. Then $(f_1,f,f_2,f,f_3,f,\dots)$ is a counter-example.
What you need is the fact that every subsequence of $(f_n)$ also has limit point in $L^{1}$. Under this assumption here is a proof:
Let $f$ be a limit point of $(f_n)$ in $L^{1}$. There is a subsequence $(f_{m_k})$ of $(f_n)$ converging to $f$ in $L^{1}$, hence also in measure. Let $f_n \to g$ in measure. Since $(f_{m_k}) \to g$ in measure we conclude that $f=g$ a.e. Thus, $\int |f_{m_k}-g| \to 0$.
This argument can be applied to any subsequence of $(f_n)$ instead of the whole sequence. Conclusion: Every subsequence of $(f_n)$ has further subsequence converging to $g$ in $L^{1}$. This implies that $\int |f_n-g| \to 0$.