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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

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  • $\begingroup$ Don’t use Math mode to fake italics. $\endgroup$ Sep 28 '20 at 21:45
  • $\begingroup$ if the limit exists then it is equal to the limsup (as well as the limif). Also $\limsup(-a_n)=-\liminf a_n$ $\endgroup$
    – alphaomega
    Sep 28 '20 at 21:46
  • $\begingroup$ @alphaomega Shouldn't the quality then be $\lim \int_X f_n - \liminf \int_{E^c} f_n = \limsup (\int_X f_n + \int_{E^c} f_n)$? $\endgroup$
    – CBBAM
    Sep 28 '20 at 21:48
  • $\begingroup$ 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$ $\endgroup$
    – alphaomega
    Sep 28 '20 at 21:55
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$-lim$ $inf\int_{E^C} f_n = lim$ $sup\int_{E^C}(-f_n)$. The equality then follows. Just check geometric intuition if unsure why this is true.

EDIT: If it makes it any easier to see it, think about making a symmetry of the sequence around the x axis. The low points become the high points and the other way around. So, the subsequence that converges to lim sup now converges to the lim inf of the inverted sequence, as it "goes down" as the original went up. As the sequence for the total integral converges, lim sup and lim inf both converge to the limit, so the "big" integral is unaffected with the change. The important thing is pulling in the minus sign

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  • $\begingroup$ 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)\$? $\endgroup$
    – CBBAM
    Sep 28 '20 at 21:57
  • $\begingroup$ 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 $\endgroup$
    – Evaristo
    Sep 28 '20 at 22:03
  • $\begingroup$ I see, thank you! $\endgroup$
    – CBBAM
    Sep 28 '20 at 23:20

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