Question 11.4: For every x ∈ [0, 1] and n ∈ N, define fn(x) to be the dista...
For every x ∈ [0, 1] and n ∈ \mathbb{N}, define f_{n}(x) to be the distance between x and the nearest point in [0, 1] of the form 10^{−n}k, where k is an integer. If f (x) = \Sigma_{n=1}^{∞}{f_{n}(x)}, evaluate \int_{[0,1]}{f} dm.
Taking x ∈ [10^{−n}k, 10^{−n}(k +1)) with 0 ≤ k ≤ 10^{n}, we have
f_{n}(x) = \begin{cases} x − 10^{−n}k, & x ∈ I_{n}^{k} \\ 10^{−n}(k + 1) − x, &x ∈ J_{n}^{k} , \end{cases}
where
I_{n}^{k} = \left[\frac{k}{10^{n}}, \frac{2k + 1}{2 × 10^{n}}\right) , J_{n}^{k} = \left[\frac{2k + 1}{2 × 10^{n}}, \frac{k + 1}{10^{n}}\right).
Using the result of Example 11.1 and the linearity of the integral, we obtain
\int_{I_{n}^{k}}{f_{n}} dm =\int_{J_{n}^{k}}{f_{n}} dm = \frac{1}{8 × 10^{2n}}
⇒ \int_{[0,1]}{f_{n}} dm = \sum\limits_{k=0}^{10^{n}−1}{\int_{I_{n}^{k}∪J_{n}^{k}}{f_{n}} dm = \frac{1}{4 × 10^{n}}} .
Since f_{n} ≥ 0, Corollary 11.3.1 gives
\int_{[0,1]}{f} dm = \sum\limits_{n=1}^{∞}{\frac{1}{4 × 10^{n}}}= \frac{1}{36}.
