Applying Fatou's Lemma to Determine the Existence of Leb. Ints.

# Applying Fatou's Lemma to Determine the Existence of Lebesgue Integrals

Recall from the Fatou's Lemma page that Fatou's lemma states that if:

• 1) $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions on $I$ that are nonnegative.
• 2) $(f_n(x))_{n=1}^{\infty}$ converges to a limit function $f$ almost everywhere on $I$.
• 3) 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 we can conclude that:

• a) $f$ is Lebesgue integrable on $I$.
• b) $\displaystyle{\int_I f(x) \: dx \leq M}$.

We will now look at a nice example of applying Fatou's lemma.

## Example 1

Use the Fatou's lemma to show that the Lebesgue integral $\displaystyle{\int_I \frac{\ln (1 - x)}{\sqrt{1 - x}}}$ exists, where $I = [0, 1]$.

We will first show that $\displaystyle{\int_I -\frac{\ln (1 - x)}{\sqrt{1 - x}}}$ exists.

We first note that the function $\displaystyle{f(x) = -\frac{\ln (1 - x)}{\sqrt{1 - x}}}$ is discontinuous at $x = 1$. Define a sequence of functions $(f_n(x))_{n=1}^{\infty}$ where $f_n$ is defined for all $n \in \mathbb{N}$ by:

(1)
\begin{align} \quad f_n(x) = \left\{\begin{matrix} -\frac{\ln (1 - x)}{\sqrt{1 - x}} & \mathrm{if} \: 0 \leq x \leq \frac{n}{n+1} \\ 0 & \mathrm{if} \: \frac{n}{n + 1} < x < 1 \end{matrix}\right. \end{align}

Notice that $\displaystyle{-\frac{\ln (1 - x)}{\sqrt{1 - x}}}$ is Lebesgue integrable on $\left [ 0, \frac{n}{n+1} \right ]$ for all $n \in \mathbb{N}$, and $0$ is Lebesgue integrable on $\left ( \frac{n}{n+1}, 1 \right )$ for all $n \in \mathbb{N}$. So by theorems presented on the Additivity of Lebesgue Integrals on Subintervals of General Intervals page we see that $f_n$ is Lebesgue integrable for all $n \in \mathbb{N}$.

It's also not hard to see that $(f_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $[0, 1]$ (everywhere except at $x = 1$). Furthermore, notice that:

(2)
\begin{align} \quad \lim_{b \to 1} \int_0^b \frac{-\ln (1 - x)}{\sqrt{1 - x}} \: dx = \lim_{b \to 1} \left [(2\ln (1 - x) - 4)\sqrt{1 - x} \right ]_1^b = \lim_{b \to 1} \left [ (2 \ln (1 - x) - 4)\sqrt{1 - x} + 4 \right ] \end{align}

Using L'Hospital's rule we see that:

(3)
\begin{align} \quad \lim_{b \to 1} (2 \ln (1 - x) - 4)\sqrt{1 - x} = \lim_{b \to 1} \frac{2 \ln (1 - x) - 4}{\frac{1}{\sqrt{1 - x}}} = \lim_{b \to 1} \frac{-2\frac{1}{1 - x}}{-\frac{1}{2}(1 - x)^{-3/2}} = \lim_{b \to 1} 4 \sqrt{1 - x} = 0 \end{align}

Therefore we see that $\displaystyle{\lim_{b \to 1} \left [ (2 \ln (1 - x) - 4)\sqrt{1 - x} + 4 \right ] = 4}$. Since each $f_n$ is nonnegative on $[0, 1]$ we se that:

(4)
\begin{align} \quad \int_0^1 f_n(x) \: dx \leq 4 \end{align}

So by Fatou's lemma we must have that $f$ is Lebesgue integrable on $I$. However, notice that:

(5)
\begin{align} \quad -f(x) = - \left ( - \frac{\ln (1 - x)}{\sqrt{1 - x}} \right ) = \frac{\ln(1 - x)}{\sqrt{1 - x}} \end{align}

Since $f$ is Lebesgue integrable on $I$ we have that $\displaystyle{-f(x) = \frac{\ln(1 - x)}{\sqrt{1 - x}}}$ is Lebesgue integrable on $I$, i.e., $\displaystyle{\int_I \frac{\ln(1 - x)}{\sqrt{1 - x}}}$ exists. The graph of $-f$ (the original function) is given below: 