📂Measure Theory

Proof of Fatou's Lemma

Theorem ¹

For a sequence $\left\{ f_{n} \right\}$ of measurable functions whose function values are non-negative, $$ \int_{E} \left( \liminf_{n \to \infty} f_{n} \right) dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$

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• $\liminf$ is the limit infimum.

Explanation

This is a lemma needed to prove the Monotone Convergence Theorem in real analysis and the Dominated Convergence Theorem. Fatou’s lemma for series, without the measurable function condition, is as follows.

Series version: For a sequence $\left\{ f_{k} : \mathbb{N} \to [0, \infty) \right\}_{k \in \mathbb{N}}$ of functions whose function values are non-negative, $$ \sum_{j=1}^{\infty} \liminf_{k \to \infty} f_{k} (j) \le \liminf_{k \to \infty} \sum_{j=1}^{\infty} f_{k} (j) \qquad , \forall j \in \mathbb{N} $$

Proof

Strategy: One needs to have a feel for simple functions. As with all lemmas, it is useful but its proof is very long and complex, so it is recommended to read it when your mental state is clear and healthy. The proof for the series form is essentially the same.

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Part 1.

$$ f : = \liminf_{n \to \infty} f_{n} \\ \displaystyle g_{n} : = \inf_{k \ge n } f_{k} $$ Letting the above, then $\displaystyle f = \lim_{n \to \infty} g_{n}$. In order to show $$ \int_{E} f dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$ it suffices to show that for every simple function $\phi \le f$, $$ \int_{E} \phi dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$ holds.

Now, for a very small positive number $\varepsilon > 0$, let us define a new simple function $$ \phi_{\varepsilon}(x) := \begin{cases} \phi (x) - \varepsilon & , \phi>0 \\ 0 & , \phi = 0 \end{cases} \ge 0 $$ Then for a sufficiently large natural number $n$, $\phi_{\varepsilon} \le g_{n} \le f$ will hold. Finally, defining $A_{k} : = \left\{ x \ | \ g_{k} \ge \phi_{\varepsilon} \right\}$, $$ A_{k} \subset A_{k+1} \\ \displaystyle \bigcup_{k=1}^{\infty} A_{k} = \mathbb{R} $$

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Part 2.

Since $\phi_{\varepsilon} \le g_{n}$ on $A_{n}$, $$ \int_{A_{n} \cap E} \phi_{\varepsilon} dm \le \int_{A_{n} \cap E} g_{n} dm $$ Since $\displaystyle g_{n} = \inf_{k \ge n } f_{k}$, for $k \ge n$, $$ \int_{A_{n} \cap E} g_{n} dm \le \int_{A_{n} \cap E} f_{k} dm $$ On the other hand, since $A_{n} \cap E \subset E$, $$ \int_{A_{n} \cap E} f_{k} dm \le \int_{E} f_{k} dm $$ Therefore, we obtain the following. $$ \int_{A_{n} \cap E} \phi_{\varepsilon} dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$

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Part 3.

Since $\phi_{\epsilon}$ was a simple function, its range can be expressed as a finite set $\left\{ c_{1} , c_{2} , \cdots , c_{r} \right\}$. Letting $B_{i} := \phi_{\epsilon}^{-1} ( \left\{ c_{i} \right\} )$, $$ \int_{A_{n} \cap E} \phi_{\varepsilon} dm = \sum_{i = 1}^{r} c_{i} m (A_{n} \cap E \cap B_{i}) $$

[7]: $A_{n} \in \mathcal{M}$, $\displaystyle A_{n} \subset A_{n+1} \implies m \left( \bigcup_{n=1}^{\infty} A_{n} \right) = \lim_{n \to \infty} m (A_{n})$

Since in Part 1 we had $A_{n} \subset A_{n+1}$ and $\displaystyle \bigcup_{n=1}^{\infty} A_{n} = \mathbb{R}$, $$ \lim_{n \to \infty} \sum_{i = 1}^{r} c_{i} m (A_{n} \cap E \cap B_{i}) = \lim_{n \to \infty} \int_{A_{n} \cap E} \phi_{\varepsilon} dm = \int_{E} \phi_{\varepsilon} dm $$ Now, combining with the result obtained in Part 2, $$ \int_{E} \phi_{\varepsilon} dm \le \liminf_{k \to \infty} \int_{E} f_{k} dm $$

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Part 4.

• Case 1. $m( \left\{ x \ | \ \phi (x) >0 \right\} ) < \infty$
  $$\displaystyle \int_{E} \phi_{\varepsilon} dm = \int_{E} \phi dm - \varepsilon m( \left\{ x \ | \ \phi (x) >0 \right\} ) $$ Taking the limit as $\varepsilon \to 0$, we obtain the following. $$ \displaystyle \int_{E} \phi_{\varepsilon} dm = \int_{E} \phi dm $$

• Case 2. $m( \left\{ x \ | \ \phi (x) >0 \right\} ) = \infty$ $$\displaystyle D_{n} : = \left\{ x \ \left| \ g(x) \ge {{1} \over {2}} \min \left\{ c_{i} \right\}_{i=1}^{n} \right. \right\} $$ Letting the above, then since $$ D_{n} \subset D_{n+1} \\ \displaystyle \bigcup_{n=1}^{\infty} D_{n} = \mathbb{R} $$ we have $\displaystyle \int_{D_{n} \cap E} g_{n} dm \to \infty$, and from the definition of $g_{n}$ we obtain the following. $$ \displaystyle \int_{D_{n} \cap E} g_{n} dm \le \int_{D_{n} \cap E} f_{k} dm \le \int_{E} f_{k} dm $$

Therefore, in either case, the following holds for every simple function $\phi \le f$. $$ \int_{E} \phi dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$

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