The Real numbers are worse than I thought

The Real number system constantly surprises me. Adding measure theory, a convenient tool that simply abstracts away book-keeping and the nastiness of the real numbers only makes the pain all the more apparent. This hit me hard as I worked through the exercises of Axler’s MIRA.

Suppose $b_1, b_2, \dotsc$ is a sequence of real numbers. Define $f: \mathbb{R} \to [0, \infty]$ by

$$\htmlId{eq-1}{f(x) = \begin{cases} \displaystyle\sum_{k=1}^\infty \frac{1}{4^k |x - b_k| } & \text{ if } x \notin \{b_1, b_2, \dotsc \}, \\ \infty & \text{ if } x \in \{b_1, b_2, \dotsc \}. \end{cases}} \tag{1}$$

Prove that $|\left\{ x \in \mathbb{R} \,:\, f(x) < 1 \right\}| = \infty$.

At first glance, it’s not even clear that there’s any point where $f(x) <1$. What if I picked $(b_k)$ to contain all rational numbers? Then for any $x$, there are infinitely many $b_k$ which are arbitrarily close to $x$! How could $f(x)$ possibly be less than one anywhere?

As is to be expected, the proof of the solution to this problem is very elegant. Create an open interval of width $2^{-k}$ around $b_k$ for every $b_k$, call each interval $I_k$. It turns out that points in those intervals are the only points where $f(x) > 1$. Moreover,

$$\htmlId{eq-2}{\mu \left( \bigcup_k I_k \right) = \sum_k \mu(I_k) = \sum_k 2^{-k} = 1,} \tag{2}$$

which essentially completes the proof. Despite the simple proof, the mental picture of what happens when the sequence of $b_k$ consists of all rational numbers still lies beyond my intuition. Clearly, this problem isn’t done for me yet.