I'm studying basic real analysis and stumbled across three big results (Fatou's lemma, Lebesgue's Dominated Convergence theorem, and the Monotone Convergence theorem) in the theory of Lebesgue integration. I've seen short and slick proofs of the LDCT and MCT from scratch (in Bogachev's Measure Theory and Bass' Graduate Real Analysis/Rudin's Real and Complex Analysis, respectively). However, I was wondering if such a proof exists for Fatou's lemma. I've seen a couple of proofs that rely on neither the MCT nor the LDCT (in particular, in Royden and Fitzpatrick's Real Analysis, and on the Wikipedia page for Fatou's lemma), and while these proofs aren't too tricky or difficult to understand, they seem considerably longer than the proofs of the other two theorems.

So to summarize: Can someone supply me with a short and slick proof, perhaps even an outline of a proof for me to work through, of Fatou's lemma that does not rely on LDCT or MCT?

Thank you!

  • 1
    $\begingroup$ Measure, Integral and Probability by Capinski and Kopp contains a proof of Fatou's lemma (theorem 4.11) that doesn't depend on Lebesgue's Dominated Convergence theorem or the Monotone Convergence theorem. However, it is an undergraduate book, so I don't know whether you will find the proof short and slick enough. $\endgroup$
    – Marc
    May 29 '14 at 19:30

Let $\lambda<1$. For any $k$, denote $$B_k := \Big\{ x \in X ~\Big|~ \forall l \geq k,~ \lambda \liminf_n f_n(x) \leq f_l(x) \Big\}.$$ Then $$ \lambda \int_{B_k} \liminf_n f_n \leq \int_{B_k} f_k \leq \int_X f_k.$$ Take $\liminf_k$ in both side : $$ \lambda \liminf_k \left( \int_{B_k} \liminf_n f_n \right) \leq \liminf_k \int_X f_k.$$

The LHS is equal to $\lambda \int_X \liminf_n f_n$ because:

  • $(B_k)_k$ is increasing.

  • $\bigcup_k B_k=X$. Indeed for any $x \in X$, either $\liminf_n f_n(x)>0$ in with case $\lambda \liminf_n f_n(x) < f_k(x)$ for $k \gg 1$, or $\liminf_n f_n(x)=0$ in which case it is trivial.

Some property about measure will tell you that $\lim_k \int_{B_k} g = \int_X g$.

  • $\begingroup$ Thank you for your response! Could you help me to see why that first inequality holds? I think that I follow the argument after that (and if I do, then this certainly fits the criteria for "snappiness!"). $\endgroup$ Jun 5 '14 at 0:39
  • 1
    $\begingroup$ Well, $f_{k}(x) = e^{-x^{2}} 1_{[-k, k]}(x)$ fails to satisfy your inequality for any $k$, since $$\liminf_{k\to\infty} f_{k}(x) = e^{-x^{2}} > 0 \quad \text{while} \quad f_{k} \equiv 0 \text{ on } [-k, k].$$ The problem is that the set of $k$ satisfying that inequality may depend on $x$. $\endgroup$ Jun 5 '14 at 2:55
  • $\begingroup$ @sos440 You are perfectly right. I'll correct the proof. $\endgroup$
    – user10676
    Jun 5 '14 at 8:55
  • $\begingroup$ Sorry to dredge up an old post, but does the fact that $\lim_{n}\int_{B_n}g=\int_Xg$ not rely on the monotone convergence theorem? Or at least some limit theorem? $\endgroup$
    – Jason
    Sep 27 '15 at 18:17
  • $\begingroup$ @Jason : sorry for late answer, I was away from stackexchange for a while. It relies on the fact that $A \mapsto \int_A g$ is a measure (which is not trivial and this fact is used to prove MCT in the usual way). I am requiring that $(B_n)_n$ is increasing, that's why I don't need any limit theorem. $\endgroup$
    – user10676
    Nov 1 '15 at 13:39

In my view Fatou's lemma and MCT are more ore less equivalent. MCT is a particular case of Fatou (with nondecreasing sequences), and ususally Fatou is proved by applying MCT for the sequence $g_n=\inf\{f_n,f_{n+1},\ldots\}$.

If you want a direct proof, and you want to understand what is going on, try the simplest case of Fatou: if $(a_n)$ and $(b_n)$ are sequences of nonnegative numbers then $$ \liminf a_n + \liminf b_n \le \liminf(a_n+b_n). $$

The general case is the same you just have a bit more technical complication.


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