Fatou's lemma requires that $f_{n}$ be non-negative. I know that there exist counterexamples such as $f_{n} = \mathbb {-1}_{[n,n+1]}$ which show that the non-negativity condition is critical. However, I don't know what role this constraint comes into play in the proof of this lemma. Where is it needed?

Here is the proof:
Let $g_{k}:=inf_{n \geq k} f_{n}$, then $\lim_{n} inf f_{n} = \uparrow \lim g_{k}$. For $n \geq k$, $\begin{align} f_{n} \geq g_{k} \Longrightarrow \int f_{n} \, d\mu \geq \int g_{k} \, d\mu \Longrightarrow \inf_{n \geq k} \int f_{n} \, d\mu \geq \int g_{k} \, d\mu \end{align} $. Therefore, $ \begin{align} \int \lim \inf f_{n} \, d\mu = \uparrow \lim_{k} \int g_{k} \, d\mu \leq \uparrow \lim_{k} \inf_{n \geq k} \int f_{n} \, d\mu = \lim_{n} \inf \int f_{n} \, d\mu \end{align} $, where in the first equation, we use Monotone Convergence Theorem.

  • 1
    $\begingroup$ I think positivity is a hypothesis in the monotone convergence theorem. $\endgroup$
    – Daniel
    Apr 16, 2021 at 4:53
  • $\begingroup$ @DanielApsley No, I think that the monotone convergence theorem only requires f to be measurable $\endgroup$
    – XXX
    Apr 16, 2021 at 5:12
  • 1
    $\begingroup$ The usual form of the monotone convergence theorem does require nonnegativity. $\endgroup$
    – PhoemueX
    Apr 16, 2021 at 5:29
  • $\begingroup$ @PhoemueX But if f is measurable, then surely $f^{+}$ and $f^{-}$ are both integrable and satisfy the condition of the theorem. $\endgroup$
    – XXX
    Apr 16, 2021 at 5:36
  • 1
    $\begingroup$ Not quite. Suppose $f_n$ is an increasing sequence. If you take $f_n^-(x)$ to be $-\min(f_n(x), 0)$, then $f_n^-$ will be a decreasing sequence. I can't seem to find a statement of the theorem that doesn't bound them below by zero. $\endgroup$
    – Daniel
    Apr 16, 2021 at 20:19

2 Answers 2


Monotone Convergence theorem requires non-negativiness. Just consider the functions $f_{n} = \mathbb {-1}_{[n,n+1]}$. Note that $f_{n} \uparrow 0$ , but $ \int f_n = 1$, for all $n$.

Of course, if $f_n$ are measurable, you can work with $f_n^+$ and $f_n^-$. The issue here is that, if $f_n \uparrow f$, then $f_n^- \downarrow f^-$. So you can not apply Monotone Convergence Theorem to the negative part (you would need Dominated Convergence Theorem).

In the proof of Fatou's lemma, that is exactly where the proof breaks if you don't assume that $f_{n}$ are non-negative:

Let $g_{k}:=\inf_{n \geq k} f_{n}$, then, for the negative part we have $g_{k}^-=\sup_{n \geq k} f_{n}^-$ and $\lim_n \sup f_n^- = \downarrow \lim g_k^-$. (then you would need Dominated Converge Theorem).

To correct this, you might be tempted to write:

Let $g_{k}:=\inf_{n \geq k} f_{n}^+$ and $h_{k}:=\inf_{n \geq k} f_{n}^-$. We have then $\lim_{n} \inf f_{n}^+ = \uparrow \lim g_{k}$ and $\lim_{n} \inf f_{n}^- = \uparrow \lim h_{k}$, but now the issue is that, although $f =f^+ - f^-$, we have that $\lim_{n} \inf f_{n} \neq \lim_{n} \inf f_{n}^+ - \lim_{n} \inf f_{n}^-$. So, although you can apply Fatou to the positive and negative parts separately, you can not combine the two results for $f$.


Monotonicity is violated

See extended proof of fatou's lemma here.

In monotonicity, we require that

  • if $X_1,X_2 \in \mathcal F$ and $X_1 \subseteq X_2$, then $\int\limits_{X_1} f d\mu \leq \int\limits_{X_2} f d\mu$. This is violated
  • if $S_1 \subseteq S_2 \subseteq S_3\subseteq \dots \subseteq S$ is a non decreasing chain of $\mu$-measurable sets, then $\int\limits_{S} f d\mu = \lim\limits_{n\to\infty} \int\limits_{S_n} f d\mu$. The proof of this requires $f$ non-negative

In your proof you need that $g_n$ is pointwise non-decreasing. But this needs positivity of $f$?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .