Show $\sum_k \frac{1}{4^k|x-b_k|}$ converges in many points. Consider the following problem from Axler's "Measure, integration and real analysis" (p72):

Honestly, I'm not even sure where to begin. This problem appears in the section that proves Lusin's and Egorov's theorem. The abstract Lebesgue integral is not yet introduced so I can not use that. Otherwise I could just show that the sum has finite integral and could conclude.
I think we can write $f= \lim_k f_k$ as a pointwise limit of the partial sums but this probably does not quite help.
Further attempt:
I tried to show $$\{x: f(x) < 1\}$$
contains a set of infinite measure but was unsuccesful.
A hint to get started is appreciated!
 A: Start with a sequence $(c_k)_{k \in \mathbb{N}}$ of (strictly) positive real numbers such that $\sum_{k = 1}^{\infty} c_k$ converges. For the exercise it may be useful to pick the sequence such that $\sum_{k = 1}^{\infty} c_k < 1$. We shall assume that this is the case, but let me point out that dropping that assumption yields further interesting results about $f$.
Then it is clear that $f(x) < 1$ at least for all $x$ such that
$$\frac{1}{4^k \lvert x - b_k\rvert} \leqslant c_k$$
holds for all $k \in \mathbb{N}\setminus \{0\}$. Taking complements we obtain
$$\{ x \in \mathbb{R} : f(x) \geqslant 1 \} \subseteq \bigcup_{k = 1}^{\infty} \{ x \in \mathbb{R} : c_k4^k\lvert x-b_k\rvert < 1\}\,.$$
Now a small rewrite shows that a suitable choice of $c_k$ proves that $\{x \in \mathbb{R} : f(x) \geqslant 1\}$ has finite measure, namely the measure is bounded by
$$\sum_{k = 1}^{\infty} \frac{2}{4^kc_k}\,.$$


Otherwise I could just show that the sum has finite integral and could conclude.

That doesn't work, the sum has infinite integral. We have
$$\int_{b_1 + \varepsilon}^{b_1 + 1} f(x)\,dx \geqslant \int_{b_1 + \varepsilon}^{b_1+1} \frac{dx}{4(x-b_1)} = \frac{1}{4}\log \frac{1}{\varepsilon}$$
for all $\varepsilon \in (0,1)$. Letting $\varepsilon \to 0$ we see that the integral is infinite.
