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

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.