# Folland’s Real Analysis, Q2.13

I have been working through this proof and came across this link. The original problem can be found in that link as well as Exercise 2. My question is in the middle of the second page they claim the following:

$$\lim \int_X f_n - \liminf \int_{E^c} f_n = \limsup (\int_X f_n - \int_{E^c} f_n)$$

I am unsure how this equality holds. Would anyone be able to explain?

Thanks

• Don’t use Math mode to fake italics. Sep 28 '20 at 21:45
• if the limit exists then it is equal to the limsup (as well as the limif). Also $\limsup(-a_n)=-\liminf a_n$ Sep 28 '20 at 21:46
• @alphaomega Shouldn't the quality then be $\lim \int_X f_n - \liminf \int_{E^c} f_n = \limsup (\int_X f_n + \int_{E^c} f_n)$? Sep 28 '20 at 21:48
• actually to be more precise it should be $\lim \int_X f_n - \liminf \int_{E^c} f_n = \lim \int_X f_n +\limsup (-\int_{E^c} f_n)\geq \limsup (\int_X f_n - \int_{E^c} f_n)$. which is enough since it is already shown that $\int_{E} f =\liminf \int_E f_n$ Sep 28 '20 at 21:55

$$-lim$$ $$inf\int_{E^C} f_n = lim$$ $$sup\int_{E^C}(-f_n)$$. The equality then follows. Just check geometric intuition if unsure why this is true.
• By pulling out the minus sign, wouldn't there be a plus sign instead? $\lim \int_X f_n - \liminf \int_{E^c} f_n = \limsup (\int_X f_n + \int_{E^c} f_n)\$? Sep 28 '20 at 21:57
• No, you are first taking the minus sign into the $\liminf$ (and thus mutating it to a $\limsup$) and then taking in the constant whole limit. But the minus sign keeps living inside the $\limsup$. It is $- \liminf\int_{E^C}f_n = \limsup( - \int_{E^C}f_n)$. Then add the constant term in Sep 28 '20 at 22:03