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: