Applying Lebesgue's Dominated Convergence Theorem 2

Applying Lebesgue's Dominated Convergence Theorem 2

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 $\displaystyle{\int_I \frac{x \ln x}{ln(x) \ln (1 + x)} \: dx}$ exists, where $I = [0, 1]$.

Notice that the function $f(x) = \ln (x) \ln (1 + x)$ is continuous on all of $I$ except for $x = 0$. Define a sequence of functions $(f_n(x))_{n=1}^{\infty}$ where $f_n$ is defined for all $n \in \mathbb{N}$ by:

\begin{align} \quad f_n(x) = \left\{\begin{matrix} 0 & 0 \leq x < \frac{1}{n}\\ \ln(x) \ln (1 + x) & \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 $\ln(x) \ln (1 + x)$ is Lebesgue integrable on $\left [ \frac{1}{n}, 1 \right ]$ for all $n \in \mathbb{N}$. So, by the theorems presented on the Additivity of Lebesgue Integrals on Subintervals of General Intervals page we have that $f_n$ is Lebesgue integrable for all $n \in \mathbb{N}$.

Moreover, it's not hard to see that $(f_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I$ (everywhere except at $x = 0$). Using L'Hospital's rule twice gives us that:

\begin{align} \quad \lim_{x \to 0} [\ln(x) \ln (1 + x)] = \lim_{x \to 0} \frac{\ln (x)}{\frac{1}{\ln(1 + x)}} = \lim_{x \to 0} \frac{\frac{1}{x}}{\frac{-\frac{1}{1 + x}}{\ln^2 (1 + x)}} = \lim_{x \to 0} \frac{-(1 + x)\ln^2 (1 + x)}{x} = \lim_{x \to 0} \left [-\ln^2 (1 + x) + -2 \ln(1 + x) \right ] = 0 \end{align}

So $f(x) = \ln (x) \ln (x+1)$ is bounded on $[0, 1]$, i.e., there exists some $M \in \mathbb{R}$, $M > 0$ such that $\mid f(x) \mid \leq M$ for all $x \in [0, 1]$. Let $g(x) = M$. Then $g$ is Lebesgue integrable on $I$. From the way each $f_n$ is defined, we also see that for all $x \in [0, 1]$ and for all $n \in \mathbb{N}$ we have that:

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

So by Lebesgue's dominated convergence theorem we can conclude that $\displaystyle{\int_I \ln (x) \ln(1 + x) \: dx}$ exists. A graph of $f$ is given below.

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