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The task would be to find a subsequence of a uniformly integrable(UI) sequence $\{ f_n\}_1^\infty$ on $[0,1]$, such that this subsequence has a limit in the $L^1[0,1]$-norm(wrt. Lebesgue measure). I already know, that this is possible in the weak sense(Dunford-Pettis theorem), and I know a counterexample, if $\{ f_n\}_1^\infty$ is only bounded in the $L^1[0,1]$-norm(here it suffices to take $f_n(x)=n\chi_{[0,\frac1n]}(x)$, $\chi$ denoting the characteristic function, but this sequence is not UI). I also don't think that it is possible in the UI-case, but I don't know an explicit counterexample.

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First note that if the statement was true, then it would also be true for complex valued sequences of functions (if $(f_n)_n$ are uniformly integrable, then so are the real and imaginary part. Now take a subsequence along which the real part converges in $L^1$, and take a further subsequence on which the imaginary part converges).

Thus, it is enough to provide a complex-valued counterexample. Since the sequence $(f_n)_n = (e^{2\pi i n x})_n$ is bounded in $L^\infty$, it is uniformly integrable. Now assume that $f_{n_k} \to f$ with convergence in $L^1$. This easily implies $\int_0^1 f \cdot e^{2 \pi i l x} d x = \lim_k \int_0^1 e^{2 \pi i (l + n_k) x} d x = 0$ for all $l \in \Bbb{Z}$. By basic Fourier analysis, this implies $f =0$ almost everywhere. Because of $\|f_n\|_{L^1} = 1$, we thus cannot have $L^1$ convergence $f_n \to f$.

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