I am reading a proof of Fatou's lemma and I don't follow a step. Consider a sequence of non-negative measurable functions on $X$ to $\overline{\mathbb{R}}$. Fatou's lemma states that

$$\int(\lim\inf f_n)\,d\mu\leq\lim\inf\int{}f_n\,d\mu$$

where $\mu$ is a measure on $X$. The proof begins by considering the sequence


therefore, if $m\leq{}n$ we have that


as long as $m\leq{}n$. The next step states that implies


I don't see this. How does this follow?

  • $\begingroup$ This follows by the definition of $\liminf$. $$\liminf_{n \to \infty} x_n = \sup \{ a : \mbox{ eventually } x_n \ge a \} $$ What definition do you have? $\endgroup$
    – Crostul
    Mar 24, 2016 at 12:29

1 Answer 1


Informally, because you might chop off the minimum when you pass to a subset. That is, we have used the fact $$ F \subset G \implies \inf F \geq \inf G \text{.} $$

The expression $\liminf \int f_n \,\mathrm{d}\mu$ asks one to compute "the lower bound of the sequence of successive tails of the sequence". At some point, that sequence of successive tails is eventually (and then forever after) a subset of $F_m = \{\int f_m, \int f_{m+1}, \dots \}$. So we subsequently only consider subsets of $F_m$, and their infima can be no less than the infimum of the full set. I.e., $$ \inf F_m \leq \inf F_{m+1} \leq \inf F_{m+2} \leq \cdots $$

  • $\begingroup$ but here you assume that in $\int{}g_m\,d\mu$ you can commute the integral and the infimum. Is this so, and if so why? $\endgroup$ Mar 29, 2016 at 12:55
  • $\begingroup$ @AnarchistBirdsWorshipFungus : Do you agree that $g_m \leq f_n \implies \int g_m \leq \int f_n$? Then, do you agree that $g_m \leq f_n, g_m \leq f_{n+1}, \dots \implies \int g_m \leq \int f_n, \int g_m \leq \int f_{n+1}, \dots$? $\endgroup$ Mar 29, 2016 at 22:59

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