Suppose that on a measure space $(X,\mathcal{M},\mu)$ I have a sequence of real-valued measurable functions $\{f_n\}_{n\geq 1}$ which is Cauchy with respect to the $L^1$ metric, i.e. $$\forall \ \varepsilon > 0, \ \exists \ N > 0 \ \text{ such that } \ \forall \ m,n > N, \ \ \int_X |f_m(x) - f_n(x)| d\mu(x) < \varepsilon.$$

Can I then conclude that all but finitely many elements of $\{f_n\}$ are integrable, i.e. elements of $L^1$?

If not, does adding the assumption that $\mu$ is $\sigma$-finite help?


  • $\begingroup$ Take $X = [0, + \infty[$ with Lebesgue measure and $f_n = id_X$, for all $n$. This is a counterexample. $\endgroup$
    – Gustavo
    Oct 14, 2015 at 4:00
  • $\begingroup$ OK, thanks @Gustavo. $\endgroup$ Oct 14, 2015 at 4:02

1 Answer 1


No. Consider Lebesgue measure on $(0,1]$ and $f_n(x) =f_0(x)=1/x$ for all $n \in N$ and for all $x\in (0,1]$ . Or if you don't want a constant series, let $f_n(x)=x^{-1}+2^{-n}.$


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