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.
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It easily follows from $|f| \leq g$ and $f_n \to f$ a.e. After this use property of liminf – Kelvin Lois Jun 02 '20 at 23:14
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See https://math.stackexchange.com/q/334114/148510. $\liminf(x_n) = - \limsup(-x_n)$ – RRL Jun 02 '20 at 23:15