Exercise 7.3.8

Question

As in Exercise 7.3.6, let $\left\{r_{1}, r_{2}, r_{3}, \ldotsight\}$ be an enumeration of the rationals in $[0,1]$, but this time define
$$
h_{n}(x)=\left\{\begin{array}{ll}
1 & 	ext { if } r_{n}<x \leq 1 \
0 & 	ext { if } 0 \leq x \leq r_{n}
\end{array} .ight.
$$
Show $H(x)=\sum_{n=1}^{\infty} h_{n}(x) / 2^{n}$ is integrable on $[0,1]$ even though it has discontinuities at every rational point.

Ulisse Mini · Accepted Answer

$H_n(x) = \sum^n_{i=1} h_i(x) / 2^n$ is integrable and $(H_n)$ uniformly approaches $H$.

Exercise 7.3.8

Answers

Comments

Exercise 7.3.8

Answers

Comments

Add answer