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 '17 at 9:33

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 '17 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 '17 at 16:12

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