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\|
Limit of an $L^1$ sequence
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real-analysis
analysis
measure-theory
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0@AD. Oh well, this is a long and ever-ongoing debate and I don't really feel like discussing this again. – 2011-05-16
2 Answers
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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 $A$).
$f_n\to f$ where $f=...$.
$\|f_n\|\le ...$ and $\|f\|=...$
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0Indeed, 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 $L^1(\mathbb{R})$.