I'm doing exercise in Real Analysis of Folland, and got stuck on this problem. I try to use Fatou lemma but can't come to the conclusion. Can anyone help me. I really appreciate.

Consider a measure space $(X, M, \mu)$.Suppose $\{f_{n}\} \in L^{+}, f_{n} \rightarrow f$ pointwise and $\int f = \lim \int f_{n} < \infty$. Then $\int_{E} f = \lim \int_{E} f_{n}$ for all $E \in M$. However, this need not be true if $\int f = \lim \int f_{n} = \infty$

Thanks so much.


By Fatou's lemma, we have

$$\int_E f \, d\mu \leq \liminf_{n \to \infty} \int_E f_n \, d\mu . \tag{1}$$


$$\int_{X \backslash E} f\, d\mu \leq \liminf_{n \to \infty} \int_{X \backslash E} f_n\, d\mu . \tag{2}$$

Since $\int_X f\, d\mu = \lim_{n \to \infty} \int_X f_n \, d\mu $, we get

$$\begin{align*} \int_{E} f\, d\mu &= \int_X f \, d\mu - \int_{X \backslash E} f\, d\mu \\ &\stackrel{(2)}{\geq} \lim_{n \to \infty} \int_X f_n \, d\mu - \liminf_{n \to \infty} \int_{X \backslash E} f_n \, d\mu \\ &= \limsup_{n \to \infty} \left( \int_X f_n \, d\mu - \int_{X \backslash E} f_n \, d\mu \right) = \limsup_{n \to \infty} \int_E f_n\, d\mu . \tag{3}\end{align*}$$

Combining $(1)$ and $(3)$ shows

$$\int_E f \, d\mu \leq \liminf_{n \to \infty} \int_E f_n \, d\mu \leq \limsup_{n \to \infty} \int_E f_n\, d\mu \leq \int_E f\, d\mu .$$

  • $\begingroup$ amazing. Very clever solution, except that I think you mistype the symbol. $X$ is the set and $M$ is algebra, so the integral in the second statement should be $\int_{X \setminus E}$, not $\int_{M \setminus E}$ $\endgroup$ Feb 16 '14 at 13:07
  • $\begingroup$ @leducquang Of course, you are right; I mixed it up. $\endgroup$
    – saz
    Feb 16 '14 at 13:23
  • $\begingroup$ I think the $= \limsup_{n \to \infty} \left( \int f_n - \int_{X \backslash E} f_n \right)$ should be $\geq \limsup_{n \to \infty} \left( \int f_n - \int_{X \backslash E} f_n \right)$. $\endgroup$ Nov 7 '15 at 16:14
  • $\begingroup$ @LeGrandDODOM Why? If $(a_n)_n$ and $(b_n)_n$ are two sequences such that $a_n$ converges, then $$\limsup (a_n + b_n ) = \lim_n a_n + \limsup_n b_n.$$ That's exactly what I used there. $\endgroup$
    – saz
    Nov 7 '15 at 16:18
  • $\begingroup$ @saz yes, my mistake. $\endgroup$ Nov 17 '15 at 16:38

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