Let $\mu$ be a $\sigma$-finite measure and $(f_n)_{n\geq1}$ and $f$ be measurable functions. Show that $f_n \rightarrow f$ in $L^1(\mu)$ iff all of the following three conditions are satisfied

  1. $f_n$ converges to $f$ in measure.

  2. The sequence $(f_n)$ is uniformly integrable, i.e. $\forall \epsilon > 0, \exists \delta > 0$ such that if $\mu(E) < δ$ then $\int_E |f_n(x)| d\mu(x) < \epsilon$ for any $n$.

  3. The sequence $(f_n)$ is uniformly tight, i.e. $\forall\epsilon > 0, \exists E, \mu(E) < \infty$ such that for any $n$ : $\int_{E^c} |f_n(x)| d\mu(x) < \epsilon$.

Remark: If $\mu$ is finite, the last condition is automatically satisfied

Remark: I am able to show the first two parts of implication, I am unable to show that $F_n$ converging in $L^1$ implies uniform tightness; also I am unable to show that given the three conditions are satisfied the sequence of measurable functions converge in $L^1$.

  • $\begingroup$ Which parts exactly can you prove? Do you mean that you're having trouble proving that $f_n$ converging in $L^1$ implies the three conditions? $\endgroup$ – Ian Oct 8 '15 at 14:14
  • $\begingroup$ no the other way round. $\endgroup$ – user275834 Oct 8 '15 at 14:26
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    $\begingroup$ The other way is a variant of the Vitali convergence theorem. Do you know what to do when you replace convergence in measure with a.e. convergence? $\endgroup$ – Ian Oct 8 '15 at 14:37
  • $\begingroup$ Thanks. Nopes, I do not not know what to do in such a case. $\endgroup$ – user275834 Oct 8 '15 at 14:40
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    $\begingroup$ To prove the uniform tightness, first take some $E$ of finite measure so that $\int_{E^c} |f(x)|d\mu(x)$ is small. Then take some $n_0$ so that for all $n\ge n_0$, $\int|f_n(x) - f(x)|d\mu(x)$ is small. Argue that then $\int_{E^c} |f_n(x)|d\mu(x)$ must be small too. Finally, if needed, "increase $E$ to let small values of $n$ in". Report if you have some problems with this approach. $\endgroup$ – zhoraster Oct 8 '15 at 15:15

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