I will first state Fatou's lemma and provide a proof, then I will present the corollary I am trying to prove. I am a little lost on the proof so I to assist in the reader's help I will provide what Folland suggests to do to prove it.

If $\{f_n\}$ is any sequence in $L^{+}$, then $$\int(\lim\inf f_n)\leq \lim\inf\int f_n$$

proof:$$\int \lim_{j\rightarrow \infty} \inf f_j = \int \lim_{k\rightarrow \infty} \inf_{k\geq j}f_k$$ by Monotone Convergence Theorem, then $$= \lim_{k\rightarrow \infty}\int \inf_{k\geq j}f_k$$ observe that $\inf_{k\geq j}f_k \leq f_k$ for all $k\geq j$. So, $$\int \inf_{k\geq j}f_j \leq \int f_k \ \ \ \forall k\geq j$$ $$\leq \inf_{k\geq j}\int f_k \leq \lim_{k\rightarrow \infty}\inf_{k\geq j}\int f_k$$

Corollary - If $\{f_n\}\subset L^{+}$, $f\in L^{+}$, and $f_n\rightarrow f$ a.e., then $\int f\leq \lim \inf \int f_n$

proof (Folland) - If $f_n\rightarrow f$ everywhere, the result is immediate from Fatou's lemma, and this can be achieved by modifying $f_n$ and $f$ on a null set without affecting the integrals by proposition 2.16 which states:

If $f\in L^{+}$, then $\int f = 0$ if and only if $f = 0$ a.e.

I am lost on how to prove the corollary I just need some initial start and I think I can go from there.


2 Answers 2


Or simpler maybe: $\lim_{n}f_{n}=\liminf_{n}f_{n}=f a.e.$ By integrating, $\int \liminf_{n}f_{n}=\int f$. By Fatou's lemma, the left integral is $\leq\liminf_{n}\int f_{n}.$ The result follows immediately.


Let $C=\{x:f_n(x)\to f(x)\}$. Modify $f_n$ and $f$ s.t. $f'=f_n'=0$ on $X\setminus C$ and $f'=f, f_n'=f_n$ on $C$. Then $f_n'\to f'$ everywhere and $f-f'=0, f_n-f_n'=0$ a.e.

$$\therefore \int f=\int f'\leq \liminf \int f_n'=\liminf \int f_n$$

  • $\begingroup$ Why did you let $f' = f_n' = 0$ on $X\setminus C$? $\endgroup$
    – Wolfy
    Nov 9, 2015 at 0:48
  • $\begingroup$ @MorganWeiss In that case $f-f'\in L^{+}$ and $\int f=\int (f-f')+\int f'=\int f'$ by proposition 2.16. $\endgroup$
    – user140541
    Nov 9, 2015 at 2:29
  • $\begingroup$ I don't think I understand, why not let $f' = f'_n = 0$ on $C$? $\endgroup$
    – Wolfy
    Nov 9, 2015 at 2:33
  • $\begingroup$ @MorganWeiss Oh. $C$ is the subset on which $f_n\to f$. So you need to modify functions on its complement... (If you set $f' = f'_n = 0$ on $C$, then $\int f'=0$) $\endgroup$
    – user140541
    Nov 9, 2015 at 2:34
  • $\begingroup$ I see, so since we are modifying the functions then we have to let $f' = f'_n = 0$ be on $X\setminus C$? $\endgroup$
    – Wolfy
    Nov 9, 2015 at 2:36

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