Fatou's Lemma

# Fatou's Lemma

From the Levi's Monotone Convergence Theorems we can deduce a very nice result commonly known as Fatou's Lemma which we state and prove below.

Lemma (Fatou's Lemma): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative Lebesgue integrable functions on an interval $I$ that converges to a function $f$ almost everywhere on an interval $I$, and suppose there exists an $M \in \mathbb{R}$, $M > 0$ such that $\displaystyle{\int_I f_n(x) \: dx \leq M}$ for all $n \in \mathbb{N}$. Then $f$ is Lebesgue integrable on $I$ and $\displaystyle{\int_I f(x) \: dx \leq M}$. |

**Proof:**Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative Lebesgue integrable functions on an interval $I$, and suppose that $(f_n(x))_{n=1}^{\infty}$ converges to a function $f$ almost everywhere on $I$. Define a new sequence of functions as follows:

\begin{align} \quad g_n(x) = \inf_{k \geq n} \{ f_n(x) \} \end{align}

- Note that $(g_n(x))_{n=1}^{\infty}$ is an increasing sequence of functions, and since $(f_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I$ we have that the following equality holds almost everywhere on $I$:

\begin{align} \quad \liminf_{n \to \infty} f_n(x) = f(x) = \limsup_{n \to \infty} f_n(x) \end{align}

- Notice that $\displaystyle{\lim_{n \to \infty} g_n(x) = \liminf_{n \to \infty} f_n(x) = f(x)}$, so $(g_n(x))_{n=1}^{\infty}$ is a nonnegative increasing sequence of functions that converges to $f$ almost everywhere on $I$.

- Now, for all $n \in \mathbb{N}$ we have that $\displaystyle{g_n(x) = \inf_{k \geq n} f_n(x) \leq f_n(x)}$ and so by comparison:

\begin{align} \quad \int_I g_n(x) \: dx \leq \int_I f_n(x) \: dx \leq M \end{align}

- Therefore $\left (\int_I g_n(x) \: dx \right )_{n=1}^{\infty}$ is an increasing sequence that is bounded above by $M$ and so it converges, i.e., $\displaystyle{\lim_{n \to \infty} \int_I g_n(x) \: dx}$ exists.

- So by the Levi's Monotone Convergence Theorems we have that since $(g_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions on $I$ that is increasing almost everywhere on $I$ and $\displaystyle{\lim_{n \to \infty} \int_I g_n(x) \: dx}$ exists, we have that $(g_n(x))_{n=1}^{\infty}$ converges almost everywhere on $I$ to some Lebesgue integrable function $g$ on $I$. But we know that $g = f$, so $f$ is Lebesgue integrable on $I$, and moreover:

\begin{align} \quad \lim_{n \to \infty} \int_I g_n(x) \: dx = \int_I \lim_{n \to \infty} g_n(x) \: dx = \int_I \lim_{n \to \infty} f(x) \: dx \leq M \quad \blacksquare \end{align}