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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!

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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. $$ \lim_n \int_X f_n d\mu = \limsup_n \int_X f_n d\mu = \liminf_n\int_X f_n d\mu=\int_X f d\mu. $$

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  • $\begingroup$ 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

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