Sheldon Axler Measure Integration Real Analysis Section 2E Exercise 14

617 Views Asked by At

As you can tell from the title, this is an exercise from Axler's measure theory book, and I am struggling with the following problem:

Suppose $b_1,b_2,\dots$ is a sequence of real numbers. Define $f: \mathbb{R} \rightarrow [0,\infty]$ by $$f(x) = \left\{\begin{array}{cc} \sum\limits_{k=1}^\infty \frac{1}{4^k|x-b_k|} & \text{ if } x \notin \{b_1,b_2,\dots\},\\ \infty & \text{ if } x \in \{b_1,b_2,\dots\} \end{array}\right.$$ Prove that $|\{x \in \mathbb{R} : f(x) < 1\}| = \infty$.

I am not sure what to do. Can someone help, please?

1

There are 1 best solutions below

0
On BEST ANSWER

Consider the sets $$ E_k = \{x : |x-b_k| > 2^{-k}\},\qquad k=1,2,3\dots,\\ E = \bigcap_{k=1}^\infty E_k . $$ If $x \in E_k$ then $$ |x-b_k| > 2^{-k} ,\\ 4^k|x-b_k| > 2^{k} ,\\ \frac{1}{4^k|x-b_k|} < 2^{-k} . $$ If $x \in E$ then $$ f(x) = \sum_{k=1}^\infty \frac{1}{4^k|x-b_k|} < \sum_{k=1}^\infty 2^{-k} = 1 . $$ What is the measure of $E$? To compute that, consider complements: $$ E_k^c = \{|x-b_k| \le 2^{-k}\},\qquad |E_k^c| = 2\cdot 2^{-k} ,\\ |E^c| = \left|\bigcup_{k=1}^\infty E_k^c\right| \le \sum_{k=1}^\infty |E_k^c| = \sum_{k=1}^\infty 2\cdot 2^{-k} < \infty , $$ and therefore $|E| = \infty$.