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We have functions $f_n\in L^1$ such that $f_n(x)$ tends to some $f(x)$ for almost all $x$. Does this mean that $f_n\to f$ in $L^1$? A necessary condition is $\|f_n\|, is it sufficient?

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    Why do you ask? What did you try?2011-05-16
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    What about $f_n = n\cdot \chi_{[0,\frac{1}{n}]}$? We have $f_{n} \to 0$ a.e. and $\|f_{n}\|_1 = 1$, so certainly we don't have $\|f_{n} - 0\|_{1} \to 0$...2011-05-16
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    @Theo Buehler: I think it is better to use hints.2011-05-16
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    Related (though the title is misleading): http://math.stackexchange.com/questions/20931/how-to-prove-that-convergence-is-equivalent-to-pointwise-convergence-in-c0-1/20934#209342011-05-16
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    @AD. Oh well, this is a long and ever-ongoing debate and I don't really feel like discussing this again.2011-05-16

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Hint:

Consider $L^1(0,1)$, look at $f_n=n\cdot\chi_{(0,1/n)}$ (here $\chi_A$ denotes the characteristic function on $A$).

  1. $f_n\to f$ where $f=...$.

  2. $\|f_n\|\le ...$ and $\|f\|=...$

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    Indeed, thanks a lot. But we seem to have the weak convergence.2011-05-16
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No. For example, if $f_n=\chi_{[n,n+1]}$ then $(f_n)_{n\ge1}$ has pointwise limit $0$ but $\|f_n\|_1=1$ for every $n$ so $f_n\not\to 0$ in $L^1(\mathbb{R})$.