I am reading this answer, and I am not sure how we get $$ \int g-\int f\leq \int g-\limsup\int f_n. $$ I see that the integral can distribute over the $-$ on the LHS, but I am not seeing how the $\liminf$ becomes a $\limsup$ on the RHS. Why is it that $$ \liminf\int(g-f_n)\leq\int g-\limsup\int f_n? $$ I am sure I am just missing something simple here.

  • $\begingroup$ It easily follows from $|f| \leq g$ and $f_n \to f$ a.e. After this use property of liminf $\endgroup$ – Si Kucing Jun 2 at 23:14
  • $\begingroup$ See math.stackexchange.com/q/334114/148510. $\liminf(x_n) = - \limsup(-x_n)$ $\endgroup$ – RRL Jun 2 at 23:15

Your Answer

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

Browse other questions tagged or ask your own question.