This one comes from Folland, Real Analysis, Problem 33 in the section titled Modes of Convergence.

Suppose $f_n \geq 0$ and $f_n \rightarrow f$ in measure, then $\int f \leq \liminf \int f_n$.

So I notice a few things first off, that since $f_n \to f$ in measure, we can find a subsequence $f_{n_j}$ which converges pointwise almost everywhere (Theorem 2.30 in Folland), and for this subsequence we may say (by Fatou's lemma using $f_n \geq 0$) that $\int f \leq \liminf \int f_{n_j}$, but it's not necessarily true that $\liminf \int f_{n_j} \leq \liminf \int f_n$, or at least I don't see how to prove it (and in general this is not true for any sequence and subsequence, while the reverse inequality is, I think).

Any tips, hints, or solutions?

  • $\begingroup$ Looking at $\lim \inf f_n$ or $\lim \sup f_n$ should help. $\endgroup$
    – N. S.
    Oct 27, 2011 at 21:38
  • 7
    $\begingroup$ Assume that the conclusion is wrong. You can find a subsequence $\{f_{n_k}\}$ such that $\liminf \int f_n d\mu =\lim_{k\to +\infty}\int f_{n_k}d\mu$. Now, extract from this subsequence an almost everywhere converging subsequence, and Fatou's lemma yields a contradiction. $\endgroup$ Oct 27, 2011 at 21:43
  • 1
    $\begingroup$ possible duplicate of Fatou's Lemma and Almost Sure Convergence (Pt. 2) $\endgroup$
    – t.b.
    Oct 27, 2011 at 22:33
  • $\begingroup$ inf can only get smaller for a larger set so your supposition that $\liminf \int f_n \leq \liminf \int f_n_j$ is true. $\endgroup$
    – user9352
    Nov 3, 2011 at 12:22

2 Answers 2


You can pass to a subsequence $f_{n_k}$ with $\int f_{n_k} \to \liminf \int f_n$ first.

This subsequence will also converge to $f$ in measure and ... then you already know what to do.

  • $\begingroup$ and....Fatou. (in other words, the comment was more helpful, but I'll mark it as an answer anyway) $\endgroup$
    – JeremyKun
    Oct 27, 2011 at 23:14
  • $\begingroup$ @Bean: Well, I didn't want to write down a complete solution, since I was under the impression that you would appreciate a hint - maybe it was not that great a hint, since it gave away too much..? (in other words, the comment came only after I had already posted this answer, but I'll gladly accept your marking my answer anyway ;) ) $\endgroup$
    – Sam
    Oct 27, 2011 at 23:37
  • $\begingroup$ Good point. the hard part was passing to the subsequence anyway, so I appreciate it. $\endgroup$
    – JeremyKun
    Oct 28, 2011 at 1:23

You can use the Urysohn subsequence principle, but it should be modified a little bit.

(Urysohn subsequence principle). Let $x_n$ be a sequence of real numbers, and let $x$ be another real number, then $\liminf x_n\geq x$ iff every subsequence $x_{n_j}$ of $x_n$ has a further subsequence $x_{n_{j_i}}$ such that $\liminf x_{n_{j_i}}\geq x$.


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