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What does it mean for a sequence $f_n$ to converge to some function, say, $f$ in the $L^1$ norm?

Is it enough to show that $\int|f_n -f| \to 0$ or must one show as well that $f\in L^1$?

I am getting confused because I've encountered questions which asked to show that $f\in L^1$ and $\lim_{n\to \infty} \int |f_n-f| =0$. Does the latter imply the former?

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up vote 6 down vote accepted

Not unless you add the condition that the sequence $f_n$ is in $L^1.$ If so, then for all $n\in \mathbb{N}$

$$ \int |f| d\mu = \int |f-f_n + f_n| d\mu \leq \int |f-f_n| d\mu + \int |f_n| d\mu$$

which is finite.

If you do not require $f_n$ to be in $L^1$ then we can easily find a sequence such that $ \int |f-f_n| d\mu \to 0 $ but $f \notin L^1.$ Just pick your favorite non-integrable function $g$ and make $f_n = f = g.$

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So if $f_n$'s are integrable, then it is enough to show that $\int|f_n-f| \to 0$?. What if the $f_n$'s are uniformly integrable? – Josh May 7 '12 at 6:29
@Josh Indeed, if the $f_n$ are integrable, then the inequality I wrote above shows that if $ \int |f-f_n| d\mu \to 0$ then $f \in L^1$ as well. – Ragib Zaman May 7 '12 at 6:31
This is kind of off topic. But does uniform integrability imply integrablity in the usual sense? – Josh May 7 '12 at 6:39
If $\mathcal{H}$ is uniformly integrable then $\mathcal{H}\subseteq L^1$ if your measure $\mu$ is finite. I do not think it is the case when $\mu$ is not finite. – Stefan Hansen May 7 '12 at 7:51
In every definition I am aware of, a uniformly integrable family is made of integrable functions, whether the measure is finite or not. – Did May 7 '12 at 11:48

The answer of Ragib Zaman is correct. However, I think that a sentence like "$\{f_n\}_n$ converges to $f$ in $L^1$" means:

  1. $f_n \in L^1$ for every $n \in \mathbb{N}$;
  2. $\lim_{n \to +\infty} \int |f_n-f| =0$.

This is a reasonable approach, since, in any (say) metric space $X$, $x_n \to x$ is rather immaterial if $x_n \notin X$. The case of a sequence $u_n = f_n -f \in L^1$ with $f_n \notin L^1$ is really a trap for homework :-)

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