Exercise 1.5.10 (Examples of uniformly integrable sequences)

• From the definition of an absolutely integrable function it follows directly that

  $$\underset{n\in \mathbb{N}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}\parallel f_n{\text{∥}}_{L^1}=\underset{n\in \mathbb{N}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}\parallel f\parallel <\infty .$$

• The constant sequence $f_n=f$ is dominated by an absolutely integrable function $f$, i.e., by itself. Considering $M\in \mathbb{N}$ in the natural domain, this allows us to use the dominated convergence theorem

  $$\underset{M\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underset{n\in \mathbb{N}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}{\int \; <!--\; nolimits\; -->}_{\{x\in X:|f(x)|\ge M\}}\left|f\right|d\mu =\underset{M\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\left|f\right|1_{\{x\in X:|f(x)|\ge M\}}=\underset{M\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\left|f\right|1_{\{x\in X:|f(x)|\ge M\}}=0=0$$

  since $\left|f\right|1_{\{x\in X:|f(x)|\ge M\}}$ converges almost everywhere pointwise to zero for obvious reasons.

• Similarly, we can use the dominated convergence to conclude

  $$\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underset{n\in \mathbb{N}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}{\int \; <!--\; nolimits\; -->}_{\{x\in X:|f(x)|\le \frac{1}{k}\}}\left|f\right|d\mu =\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\left|f\right|1_{\{x\in X:|f(x)|\le \frac{1}{k}\}}=\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\left|f\right|1_{\{x\in X:|f(x)|\le \frac{1}{k}\}}=0=0$$

  since $\left|f\right|1_{\{x\in X:|f(x)|\le \frac{1}{k}\}}$ converges almost everywhere pointwise to zero for obvious reasons.

Then it is easy to verify that ${\left(f\right)}_{n\in \mathbb{N}}$ is uniformly integrable, yet any function $g$ dominating ${\left(f\right)}_{n\in \mathbb{N}}$ must have an infinite integral.