Fatou's Lemma for Nonnegative Measurable Functions

# Fatou's Lemma for Nonnegative Measurable Functions

Recall from the Fatou's Lemma for Nonnegative Lebesgue Measurable Functions page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$ such that:

• 1) $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ pointwise of $E$.

Then:

(1)
\begin{align} \quad \int_E f \leq \liminf_{n \to \infty} \int_E f_n \end{align}

We now prove Fatou's Lemma for general complete measure spaces.

 Lemma 1 (Fatou's Lemma for General Measure Spaces): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative measurable functions defined on a Lebesgue measurable set $E$. If: 1) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$. Then $\displaystyle{\int_E f(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu}$.
• Proof: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative measurable functions defined on a Lebesgue measurable set $E$ that converges pointwise to $f(x)$ on $E$.
• Let $\varphi$ be a simple function on $E$ such that $0 \leq \varphi(x) \leq f(x)$. We want to show that:
(2)
\begin{align} \quad \int_E \varphi(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}
• We will then take the supremum of the lefthand side for the conclusion of Fatou's lemma. There are two cases to consider.
• Case 1: Suppose that $\displaystyle{\int_E \varphi(x) \: d \mu = \infty}$. Since $\varphi(x)$ is a simple function, for $a_1, a_2, ..., a_n > 0$ as the positive values in the range of $\varphi$ where $\varphi(x) = a_k$ if and only if $x \in E_k \subseteq E$, we have that $\displaystyle{\int_E \varphi(x) \: d \mu = \sum_{k=1}^{n} a_k \mu (E_k)}$. So $\displaystyle{\sum_{k=1}^{n} a_k \mu (E_k) = \infty}$. But this is a finite sum. So there exists an $i_0 \in \{ 1, 2, ..., n \}$ such that $a_{i_0} \mu (E_{i_0}) = \infty$. So $\mu (E_{i_0}) = \infty$. Let $a^* = \frac{a_{i_0}}{2}$ and $E^* = E_{i_0}$.
• For each $n \in \mathbb{N}$ let:
(3)
\begin{align} \quad A_n = \left \{ x \in E : f_k(x) \geq a^*, \: \mathrm{for \: all \:} k \geq n \right \} \end{align}
• Note that for each $n \in \mathbb{N}$, $A_n = \left \{ x \in E : f_k(x) \geq a^*, \: \mathrm{for \: all \:} k \geq n \right \} \subseteq \left \{ x \in E : f_k(x) \geq a^*, \: \mathrm{for \: all \:} k \geq n + 1 \right \} = A_{n+1}$. Let $x \in E^*$. Then $\varphi(x) = a^*$. Since $(f_n(x))_{n=1}^{\infty}$ is converges to $f(x)$ on $E$ and $0 \leq \varphi(x) \leq f(x)$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $f_n(x) > a^*$. So $x \in A_{N}$. This shows that:
(4)
\begin{align} \quad E^* \subseteq \bigcup_{n=1}^{\infty} A_n \end{align}
• By the continuity and monotone properties of the measure $\mu$ we have that:
(5)
\begin{align} \quad \infty = \mu (E^*) \leq \mu \left ( \bigcup_{n=1}^{\infty} A_n \right ) = \lim_{n \to \infty} \mu (A_n) \end{align}
• For each $n \in \mathbb{N}$ sufficiently large we have that:
(6)
\begin{align} \quad a^* \mu (A_n) = \int_{A_n} a^* \leq \int_{A_n} f_n(x) \: d \mu \leq \int_E f_n(x) \: d \mu \end{align}
• We take the limit inferior of both sides of the equation above to get:
(7)
\begin{align} \quad \liminf_{n \to \infty} a^* \mu (A_n) \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \\ \quad \infty \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}
• Therefore:
(8)
\begin{align} \quad \int_E \varphi(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}
• Since $\varphi$ is arbitrary we conclude that:
(9)
\begin{align} \quad \int_E f(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}
• Case 2: Suppose that $\displaystyle{\int_E \varphi(x) \: d \mu < \infty}$. Let:
(10)
\begin{align} \quad A = \{ x \in E : \varphi(x) > 0 \} \end{align}
• Since $\varphi$ is a measurable function, this set is a measurable set. Furthermore, if $a_1, a_2, ..., a_n > 0$ are the positive numbers in the range of $\varphi$ such that $\varphi(x) = a_k$ if and only if $x \in E_k \subseteq E$ then we have that:
(11)
\begin{align} \quad A = \bigcup_{k=1}^{n} E_k \end{align}
• Note that $\mu (A) < \infty$ otherwise the integral $\displaystyle{\int_E \varphi(x) \: d \mu}$ would not be finite. Let:
(12)
\begin{align} \quad M = \max \{ a_1, a_2, ..., a_n \} \end{align}
• Let $\epsilon$ be given with $0 < \epsilon < 1$. For each $n \in \mathbb{N}$ let:
(13)
\begin{align} \quad A_n = \{ x \in E : f_k(x) > (1 - \epsilon) \varphi(x), \: \mathrm{for \: all \:} k \geq n \} \end{align}
• Once again, for all $n \in \mathbb{N}$ we see that $A_n = \{ x \in E : f_k(x) > (1 - \epsilon) \varphi(x), \: \mathrm{for \: all \:} k \geq n \} \subseteq \{ x \in E : f_k(x) > (1 - \epsilon) \varphi(x), \: \mathrm{for \: all \:} k \geq n+1 \} = A_{n+1}$.
• Furthermore, if $x \in A$ then $0 < \varphi(x)$. But we know that $0 \leq \varphi(x) \leq f(x)$ and $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$. So there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $(1 - \epsilon) \varphi(x) \leq f_n(x)$. So $x \in A_N$, and:
(14)
\begin{align} \quad A \subseteq \bigcup_{n=1}^{\infty} A_n \end{align}
• Now since $A_n \subseteq A_{n+1}$ for each $n \in \mathbb{N}$ we have that $A \setminus A_n \supseteq A \setminus A_{n+1}$ for each $n \in \mathbb{N}$.
• By the continuity properties of measure we have that:
(15)
\begin{align} \quad 0 \ mu ( \emptyset ) = \mu \left ( A \setminus \bigcup_{n=1}^{\infty} A_n \right ) = \mu \left ( A \cap \bigcap_{n=1}^{\infty} A_n^c \right ) = \mu \left ( \bigcap_{n=1}^{\infty} A \cap A_n^c \right ) = \mu \left ( \bigcap_{n=1}^{\infty} (A \setminus A_n) \right ) = \lim_{n \to \infty} \mu (A_n) \end{align}
• So for the given $\epsilon$ with $0 < \epsilon < 1$ there exists an $N^* \in \mathbb{N}$ such that for all $n \geq N^*$ we have that:
(16)
\begin{align} \quad \mu (A \setminus A_n) < \epsilon \end{align}
• Therefore, for all $n \geq N^*$ we have that:
(17)
\begin{align} \quad \int_E f_n(x) \: d \mu \geq \int_{A_n} f_n(x) \: d \mu \geq \int_{A_n} (1 - \epsilon) \varphi(x) \: d \mu \\ & \geq (1 - \epsilon) \int_{A_n} \varphi(x) \: d \mu \\ & \geq (1 - \epsilon) \left [ \int_E \varphi(x) \: d \mu - \int_{E \setminus A_n} \varphi(x) \: d \mu \right ] \\ & \geq (1 - \epsilon) \left [ \int_E \varphi(x) \: d \mu - \int_{A \setminus A_n} \varphi(x) \: d \mu \right ] \\ & \geq (1 - \epsilon) \left [\int_E \varphi(x) \: d \mu - \int_{A \setminus A_n} M \: d \mu \right ] \quad \left ( \mathrm{since} \: \int_{A \setminus A_n} \varphi(x) \: d \mu \leq \int_{A \setminus A_n} M \: d \mu \right ) \\ & \geq (1 - \epsilon) \left [ \int_E \varphi(x) \: d \mu - M \mu (A \setminus A_n) \right ] \\ & > (1 - \epsilon) \left [ \int_E \varphi(x) \: d \mu - M \epsilon \right ] \\ & > (1 - \epsilon) \int_E \varphi(x) \: d \mu - M \epsilon \\ & > \int_E \varphi(x) - \epsilon \left [ \int_E \varphi(x) \: d \mu + M \right ] \end{align}
• Taking the limit inferior of both sides of the inequality above yields:
(18)
\begin{align} \quad \liminf_{n \to \infty} \left [\int_E \varphi(x) - \epsilon \left [ \int_E \varphi(x) \: d \mu + M \right ] \right ] \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \\ \quad \int_E \varphi(x) - \epsilon \left [ \int_E \varphi(x) \: d \mu + M \right ] \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}

Since this holds for all $\epsilon$ with $0 < \epsilon < 1$, by letting $\epsilon \to 0$ we get:

(19)
\begin{align} \quad \int_E \varphi(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \quad \blacksquare \end{align}
• Since $\varphi$ is arbitrary we conclude that:
(20)
\begin{align} \quad \int_E f(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \quad \blacksquare \end{align}