How can I proof the general Fatou's Lemma without using the Monotone convergence Theorem.


Let $(X,\mathcal{M},\mu)$ be a measure space and $\{f_n\}$ a non-negative measurable sequence. Then

$$ \int_X \liminf_{n\to\infty}f_n~d\mu \leq \liminf_{n\to\infty}\int_X f_n~d\mu.$$

  • $\begingroup$ By truncation in concert with the bounded convergence theorem (assuming $\mu$ is $\sigma$-finite). $\endgroup$
    – cardinal
    Mar 31, 2012 at 21:10
  • $\begingroup$ Also, the $\liminf$ should be on the inside of the integral on the left. :) $\endgroup$
    – cardinal
    Mar 31, 2012 at 21:12
  • $\begingroup$ Why would you want to avoid the use of the monotone convergence theorem? It follows very directly from the definition of the integral. It isn't as if you would avoid a lot of complexity by not using it. $\endgroup$ Mar 31, 2012 at 21:58
  • $\begingroup$ @NateEldredge: I wouldn't avoid using MCT. But I was taught Fatou's lemma first, and that was used to prove the MCT and I was wondering if there was another way. Obviously, using the MCT to prove the general Fatou is quite easy...:) $\endgroup$
    – Kuku
    Mar 31, 2012 at 22:11

1 Answer 1


Here is one proof based on the bounded convergence theorem, adapted from Durrett.

Define $g_n(x) = \inf_{m\geq n} f_m(x)$. So, $f_n \geq g_n$ and $g_n \uparrow g(x) := \liminf_n f_n(x)$ as $n \to \infty$.

By monotonicity of the integral, we know that $\newcommand{\du}{\,\mathrm d \mu} \int f_n \du \geq \int g_n \du$, whence $$\liminf_n \int f_n \du \geq \liminf_n \int g_n \du \>.$$

Suppose $X_n \uparrow X$ where $\mu(X_n) < \infty$. By the bounded convergence theorem, for fixed $m$, we have $$ \liminf_n \int g_n \du \geq \int_{X_m} g_n \wedge m \du \to \int_{X_m} g \wedge m \du \>, $$ since the integrand in the middle is bounded and converges to the integrand on the right.

But, then $$ \liminf_n \int g_n \du \geq \sup_m \int_{X_m} g \wedge m \du = \int \liminf_n f_n \du \>. $$

Since $\liminf_n \int f_n \du \geq \liminf_n \int g_n \du$, we are done.

  • $\begingroup$ Thanks for your answer. I'm not familiar with $\wedge$. $\endgroup$
    – Kuku
    Mar 31, 2012 at 21:40
  • $\begingroup$ $x \wedge y = \min(x,y)$. $\endgroup$
    – cardinal
    Mar 31, 2012 at 21:40
  • $\begingroup$ oh ok. thanks very much. $\endgroup$
    – Kuku
    Mar 31, 2012 at 21:44

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