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Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable. In the literature, assuming the measure space $X$ has finite measure, if $f_n$ converges to $f$ in $L^{\infty}$-norm , then $f_n$ converges to $f$ in $L^{1}$-norm.

Even if $X$ has infinite measure, does it converge to $f$ in $L^{1}$-norm?

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No. Try $f_n$ the constant function such that $f_n(x)=\frac1n$ for every $x$ in $X$.

For an example where each $g_n$ is in $L^1\cap L^\infty$, $g_n\to0$ in $L^\infty$ and not in $L^1$, try $g_n=\frac1n\mathbf 1_{[0,n^2]}$ on $X=\mathbb R$ with Lebesgue measure.

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  • $\begingroup$ Here the $f_n$ are not even in $L^1$. Maybe the question can be adjusted: if $f_n\in L^1\cap L^\infty$ and $f_n$ converges in $L^\infty$, does it converge in $L^1$? $\endgroup$ – Hagen von Eitzen Oct 3 '12 at 13:11
  • $\begingroup$ @HagenvonEitzen See Edit. $\endgroup$ – Did Oct 3 '12 at 13:19

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