Fatou's Lemma for Nonnegative Lebesgue Measurable Functions

# Fatou's Lemma for Nonnegative Lebesgue Measurable Functions

Lemma 1 (Fatou's Lemma): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$. Suppose that:1) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ almost everywhere on $E$.Then $\displaystyle{\int_E f \leq \liminf_{n \to \infty} \int_E f_n}$. |

**Proof:**We may assume that $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ on all of $E$ since equality of integrals almost everywhere on a set of measure zero results in equality of integrals.

- Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$ such that $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on all of $E$.

- Let $\varphi$ be a bounded Lebesgue measurable function such that $0 \leq \varphi(x) \leq f(x)$ on $E$ and such that $\varphi$ vanishes outside a set of measure $0$. Since $\varphi$ is bounded on $E$ there exists an $M \in \mathbb{R}$, $M > 0$ such that $0 \leq \varphi(x) \leq M$ for all $x \in E$. For each $n \in \mathbb{N}$ define:

\begin{align} \quad \varphi_n(x) = \min \{ \varphi(x), f_n(x) \} \end{align}

- Then for all $n \in \mathbb{N}$ and for all $x \in E$ we have that each $\varphi_n$ is bounded on $E$ and that:

\begin{align} \quad 0 \leq \varphi_n(x) \leq \varphi(x) \leq M \end{align}

- Since $\varphi$ vanishes outside a set of finite measure there exists a subset $E_0 \subseteq E$ such that $m(\{ \varphi(x) : \varphi(x) > 0 \}) < \infty$ and $\varphi(x) = 0$ for all $x \in E \setminus E_0$.

- For the set of $x \in E$ we have that:

\begin{align} \quad \lim_{n \to \infty} \varphi_n(x) = \lim_{n \to \infty} \min \{ \varphi(x), f_n(x) \} = \min \{ \varphi(x), f(x) \} = \varphi(x) \end{align}

- For the set of $x \in E \setminus E_0$ we have that:

\begin{align} \quad \int_{E \setminus E_0} \varphi_n(x) = \int_{E \setminus E_0} 0 = 0 \end{align}

- From above we get that:

\begin{align} \quad \int_E \varphi_n &= \int_{E_0} \varphi_n + \int_{E \setminus E_0} \varphi_n = \int_{E_0} \varphi_n + 0 = \int_{E_0} \varphi_n \quad (*) \end{align}

- And we also get that:

\begin{align} \quad \int_E \varphi &= \int_{E_0} \varphi + \int_{E \setminus E_0} \varphi = \int_{E_0} \varphi + 0 = \int_{E_0} \varphi \quad (**) \end{align}

- So $(\varphi_n(x))_{n=1}^{\infty}$ is a uniformly bounded sequence of Lebesgue measurable functions defined on a Lebesgue measurable set $E_0$ with $m(E_0) < \infty$, and $(\varphi_n(x))_{n=1}^{\infty}$ converges pointwise to $\varphi(x)$ on $E_0$. So by The Uniform Bounded Convergence Theorem for Pointwise Convergent Sequences of Functions we have that:

\begin{align} \quad \lim_{n \to \infty} \int_{E_0} \varphi_n = \int_{E_0} \varphi \quad (***) \end{align}

- We now use $(*)$, $(**)$, and $(***)$ to conclude that:

\begin{align} \quad \int_E \varphi \overset{(*)} = \int_{E_0} \varphi \overset{(***)} = \lim_{n \to \infty} \int_{E_0} \varphi_n \overset{(*)} = \lim_{n \to \infty} \int_E \varphi_n = \liminf_{n \to \infty} \int_E \varphi_n \leq \liminf_{n \to \infty} \int_E f_n \quad \blacksquare \end{align}

- Therefore:

\begin{align} \quad \int_E f &= \inf \left \{ \int_{E_{\varphi}} \varphi : \varphi \: \mathrm{is \: bounded \: and \: Lebesgue \: measurable}, 0 \leq \varphi(x) \leq f(x) \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{vanishes \: outside \: of \:} E_{\varphi} \right \} \\ & \leq \liminf_{n \to \infty} \int_E f_n \end{align}