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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 $\|f_n\|<M$, is it sufficient?

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Why do you ask? What did you try? – AD. May 16 '11 at 7:07
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 $\|f_{n} - 0\|_{1} \to 0$... – t.b. May 16 '11 at 7:08
@Theo Buehler: I think it is better to use hints. – AD. May 16 '11 at 7:14
Related (though the title is misleading):… – Shai Covo May 16 '11 at 7:14
@AD. Oh well, this is a long and ever-ongoing debate and I don't really feel like discussing this again. – t.b. May 16 '11 at 7:20
up vote 1 down vote accepted


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. – peno May 16 '11 at 7:15

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 $L^1(\mathbb{R})$.

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Thank you very much. – peno May 16 '11 at 7:21

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