Applying Lebesgue's Dominated Convergence Theorem 1

Applying Lebesgue's Dominated Convergence Theorem 1

Recall from the Lebesgue's Dominated Convergence Theorem page that if:

  • 1) $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions on $I$.
  • 2) $(f_n(x))_{n=1}^{\infty}$ converges to a limit function $f$ almost everywhere on $I$.
  • 3) There exists a Lebesgue integrable function $g$ on $I$ such that $\mid f_n(x) \mid \leq g(x)$ almost everywhere on $I$ and for all $n \in \mathbb{N}$ (dominating criterion).

Then we can conclude that:

  • a) $f$ is Lebesgue integrable on $I$.
  • b) $\displaystyle{\int_I f(x) \: dx = \int_I \lim_{n \to \infty} f_n(x) \: dx = \lim_{n \to \infty} \int_I f_n(x) \: dx}$.

We will now look at applying Lebesgue's dominated convergence theorem.

Example 1

Use Lebesgue's dominated convergence theorem to show that the Lebesgue integral $\displaystyle{\int_I \frac{x \ln x}{(1 + x)^2} \: dx}$ exists, where $I = [0, 1]$.

Notice that on $I$ we have that $\displaystyle{f(x) = \frac{x \ln x}{(1 + x)^2}}$ is undefined since $\ln 0$ is undefined. Now, consider the following sequence $(f_n(x))_{n=1}^{\infty}$ of functions where $f_n$ is defined for all $n \in \mathbb{N}$ by:

\begin{align} \quad f_n(x) = \left\{\begin{matrix} 0 & \mathrm{if} \: 0 \leq x < \frac{1}{n} \\ \frac{x \ln x}{(1 + x)^2} & \mathrm{if} \: \frac{1}{n} \leq x \leq 1 \end{matrix}\right. \end{align}

Notice that $0$ is Lebesgue integrable on $\left [0, \frac{1}{n} \right )$ (for all $n \in \mathbb{N}$), and moreover, $\displaystyle{\frac{x \ln x}{(1 + x)^2}}$ is Lebesgue integrable on $\left [ \frac{1}{n}, 1 \right ]$ since $\displaystyle{\frac{x \ln x}{(1 + x)^2}}$ is continuous on every closed and bounded interval $\left [ \frac{1}{n}, 1 \right ]$ (for each $n \in \mathbb{N}$). So, from the theorems presented on the Additivity of Lebesgue Integrals on Subintervals of General Intervals page we see that $f_n$ is Lebesgue integrable on $I = [0, 1]$ for all $n \in \mathbb{N}$.

Furthermore, it's not hard to see that as $(f_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I$.

Now notice that for all $x \in (0, 1]$ that:

\begin{align} \quad \biggr \rvert \frac{x \ln x}{(1 + x)^2} \biggr \rvert \leq \mid x \ln x \mid \leq 1 \end{align}

To show this, notice that if $h(x) = x \ln x$ then $h'(x) = \ln x + 1$. So $h'(x) = 0$ when $\ln x + 1 = 0$, i.e., when $\ln x = - 1$, i.e., $x = \frac{1}{e}$. It's not hard to check that $h$ has a local minimum at $x = \frac{1}{e}$. So $h(e^{-1}) = -e^{-1}$ and $\mid h(e^{-1}) \mid \leq 1$.

We must now check the endpoints of $h$ on this interval. Notice that $h(1) = 0 \leq 1$. Also, by using L'Hospital's rule, we see that:

\begin{align} \quad \lim_{x \to \infty} x \ln x = \lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = 0 \end{align}

So $\mid h(x) \mid = \mid x \ln x \mid \leq 1$ for all $x \in (0, 1]$. So let $g(x) = 1$. Then $g$ is clearly Lebesgue integrable on $[0, 1]$, and the following equality holds almost everywhere for all $n \in \mathbb{N}$:

\begin{align} \quad \mid f_n(x) \mid \leq g(x) \end{align}

So, by Lebesgue's dominated convergence theorem we can conclude that $\displaystyle{f(x) = \frac{x \ln x}{(1 + x)^2}}$ is Lebesgue integrable on $I$, i.e., the Lebesgue integrable $\displaystyle{\int_I \frac{x \ln x}{(1 + x)^2} \: dx}$ exists for $I = [0, 1]$. The graph of $f$ is given below.

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