The question was:

Given $\mu$ a positive measure in $(X, \Sigma)$ and $f_n, f:X\rightarrow [0,\infty)$ $\mu$-summable then show that if

$\liminf f_n\geq f$ almost everywhere and $$\limsup_n \int_X f_nd\mu \leq \int_X fd\mu$$ then $f_n\to f$ in $ L^1$.

The hint was to prove that $g_n=\inf_{k\geq n} f_k$ satisfies $g_n\to f$ in $L^1$> I have done that via Fatou's Lemma and monotone convergence theorem, but I could not infer the main result!


Fatou's lemma yields that $$ \int_X \liminf_n f_n d\mu \leq \liminf _n\int_X f_n d\mu $$ and since $f\leq \liminf_n f_n$ a.e. we have that $$ \int_X f d\mu \leq \liminf_n \int_X f_n d\mu $$ and hence $$ \limsup_n \int_X f_n d\mu \leq \int_X f d\mu \leq \liminf_n\int_X f_n d\mu. $$ But as $\liminf_n a_n\leq \limsup a_n$ for any sequence $(a_n)$ the limit exists and we must have equality, i.e. $$ \lim_n \int_X f_n d\mu = \limsup_n \int_X f_n d\mu = \liminf_n\int_X f_n d\mu=\int_X f d\mu. $$

  • $\begingroup$ Oh, sorry I just realized that we want convergence in $L^1$ and not just convergence of the integrals. Nevermind my post for now. $\endgroup$ – Stefan Hansen Mar 23 '12 at 13:51
  • $\begingroup$ Using a similar argument we have that $g_n\to f$ in L^1 but we need convergence in L^1 for $f$. $\endgroup$ – checkmath Mar 23 '12 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.