# Uniformly integrable sequence has norm convergent subsequence?

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.

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 $$f_n \to f$$.