Theorem: (Fatou's lemma). Let $(f_n)_{n\geq 1}$ be a sequence of measurable functions with values in $[0,\infty]$. Then $$\int_S\liminf\limits_{n\to\infty}f_nd\mu \leq \liminf\limits_{n\to\infty}\int_S f_n d\mu.$$

Exercise: Assume that $f, f_1, f_2, ...:S\to\overline{\mathbb{R}}$ are measurable functions such that

i) There is a constant $M\geq 0$ such that for all $n\in\mathbb{N}, \int_S\left|f_n\right|d\mu\leq M$.

ii) $f_n\to f$ pointwise.

Use Fatou's lemma to show that $\int_S\left|f\right|d\mu\leq M$.

Solution: Fatou's lemma yields that: $$\int_S\left|f\right|d\mu = \int_S\liminf\limits_{n\to\infty}\left|f_n\right|d\mu\leq \liminf\limits_{n\to\infty}\int_S f_nd\mu \leq M.$$

My question: How do we know that $\int_S\left|f\right|d\mu = \int_S\liminf\limits_{n\to\infty}\left|f_n\right|d\mu$ and $\liminf\limits_{n\to\infty}\int_S f_nd\mu \leq M$?

Thanks in advance!

  • $\begingroup$ $f_n\to f$ pointwise implies that $|f_n|\to |f|$ pointwise and consequently $\liminf|f_n|=|f|$ $\endgroup$
    – drhab
    Dec 21, 2017 at 9:33

1 Answer 1


1) If $f_n \to f$ pointwise, then also $|f_n| \to |f|$ pointwise, so that $$ |f(x)| = \lim_n |f_n(x)| = \liminf_n |f_n(x)| \qquad\forall x\in S. $$

2) In general, if $(a_n)\subset\mathbb{R}$ is a sequence such that $a_n \leq M$ for every $N$, then $\liminf_n a_n \leq M$ (and also $\limsup_n a_n \leq M$).

Since, by assumption, $a_n := \int_S |f_n| \, d\mu\leq M$ for every $n$, you have that $$ \liminf_n a_n = \liminf_n \int_S |f_n|\, d\mu \leq M. $$

  • $\begingroup$ Thanks for your reply! Could you elaborate on why $f_n\to f$ pointwise implies that $\int_S \left| f\right | d\mu = \int_S\liminf\limits_{n\to\infty}\left|f_n\right|d\mu$? $\endgroup$
    – titusAdam
    Dec 22, 2017 at 11:39
  • $\begingroup$ Because $f_n\to f$ pointwise implies that $\lvert f_n\rvert\to \lvert f\rvert$ pointwise, which implies that for any $x$, $$\liminf_{n\to \infty} \lvert f_n(x)\rvert = \lim_{n\to \infty} \lvert f_n(x)\rvert = \lvert f(x)\rvert$$ We typically write this as $$\liminf_{n\to \infty} \lvert f_n\rvert = \lvert f\rvert$$ since the two are equivalent as functions. $\endgroup$
    – Michael L.
    Dec 22, 2017 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.