I am reading Royden's proof on Fatou's lemma.

Let ${f_n}$ be a sequence of nonnegative measurable functions on $E$ converging to $f$ pointwise almost everywhere on $E$, then $\int_E f \leq \liminf\int_E f_n$.

In the very last line of Royden's proof, it says

By the definition of the integral of $f_n$ over $E$, $\int_E h_n \leq \int_E f_n$ (where $h_n = \min (h , f_n)$ and $h$ is a bounded measurable function with finite support with $h \leq f$) Thus $$\int_E h = \lim \int_E h_n \leq \liminf \int_E f_n.$$

And I don't see how the last statement is implied, how does he apply the first line to imply the last? Just because $\int_E h_n \leq \int_E f_n$ doesn't tell me that $\inf \int_E f_n$ has to be greater than any particular $\int h_n$. How do we know that $\int_E f_{n+1}$ isn't smaller than $\int_E h_n$?

  • $\begingroup$ and what is $h_n$? $\endgroup$ May 13, 2018 at 12:56
  • $\begingroup$ added in edit, thanks for reminding me of the lack of clarity! $\endgroup$ May 13, 2018 at 12:58
  • $\begingroup$ ok i think it's just because $lim \int_E f_n = lim inf \int_E f_n$? But I am uneasy about the case where the $lim \int_E f_n$ does not converge $\endgroup$ May 13, 2018 at 13:06
  • $\begingroup$ I've edited your question to improve the formatting. Please use > like <blockquotes> in HTML for a direct cite. $\endgroup$ May 13, 2018 at 13:11

1 Answer 1


You can see this in two steps: Since $$ \int_E h_n \; \mathrm dx \leq \int_E f_n \; \mathrm dx \quad \text{for all } n \in \mathbb N $$ we get $$ \int_E h \; \mathrm dx = \lim_{n \to \infty} \int_E h_n \; \mathrm dx \leq \int_E f_n \; \mathrm dx \quad \text{for all } n \in \mathbb N \; . $$ By taking the $\liminf$ on the right-hand side, we finally get $$ \int_E h \; \mathrm dx \leq \sup_{n \geq 0} \inf_{m \geq n} \int_E f_m \; \mathrm dx = \liminf_{n \to \infty} \int_E f_n \; \mathrm dx \; . $$

Edit: The above explanation is not fully correct. Just take the $\liminf$ on both sides of line 1. Then we get $$ \int_E h \; \mathrm dx = \lim_{n \to \infty} \int_E h_n \; \mathrm dx = \liminf_{n \to \infty} \int_E h_n \; \mathrm dx \leq \liminf \int_E f_n \; \mathrm dx \; . $$

  • $\begingroup$ Shouldn't we also have a lim in front of $\int_E f_n$ on the second line? Since $h_n$ approaches to $h$, but there might still be some $f_n$ that are less than $h$ for n sufficiently small? $\endgroup$ May 13, 2018 at 13:19
  • 1
    $\begingroup$ @Ecotistician: You're right, the second line only holds for sufficiently large $n$. Maybe it's easier to see the inequality by taking the $\liminf$ on both sides of line 1: On the left-hand side we then have $\liminf_{n \to \infty} \int_E h_n \; \mathrm dx = \lim_{n \to \infty} \int_E h_n \; \mathrm dx$. $\endgroup$
    – aexl
    May 13, 2018 at 14:09
  • $\begingroup$ okay I'm convinced your statement is true in the case that the integrals of hn and fn converges, but what about the case that $\int_E f_n$ doesm't converge? Just thinking about it in terms of a limit of a sequence, Couldn't I have the integrals vary between two values with one less than $\int_E h$ and the other greater than it infinitely? $\endgroup$ May 13, 2018 at 19:47
  • $\begingroup$ I guess I just need to convince myself that if $a_n \leq b_n$ then $ lim inf a_n \leq lim inf b_n$. $\endgroup$ May 13, 2018 at 20:33
  • $\begingroup$ @Ecotistician: Exactly, that's the point. $\endgroup$
    – aexl
    May 13, 2018 at 23:32

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