1
$\begingroup$

I'm facing this problem,

Let $g:[a,b]\rightarrow \mathbb{R}$ be Riemann-integrable, $f:[a,b]\rightarrow \mathbb{R}$ a bounded function, $(x_n)$ a sequence of points in $[a,b]$ such that $f(x)=g(x)$ for all $x$ in $[a,b]$ other than the $x_n$. $\textbf{Give an example}$ to show that $f$ need not be Riemann-integrable.

Before this, the book ("A First Course in Real Analysis by Sterling Berberian",Page 164) says that for $f:[a,b]\rightarrow \mathbb{R}$ and $g:[a,b]\rightarrow \mathbb{R}$ Riemann integrable, and $f=g$ almost everywhere we can conclude that $f$ is Riemann-integrable.

Checking the examples they give on non riemann integrability, I found $f(x):[0,1]\rightarrow \mathbb{R}$, $f(x)=1$ for rationals and $f(x)=0$ for irrationals. From this I can build a function $g:[0,1]\rightarrow \mathbb{R}$, that can be equal to $f$ over the irrationals, and something different than $f$ for rationals, However, how could I build a sequence such as the one they ask me to, but over the rationals?

$\endgroup$
  • $\begingroup$ $\mathbb Q$ is countable and you can write $\mathbb Q$ as a sequence $\endgroup$ – user99914 Jul 13 '18 at 4:57
1
$\begingroup$

You can list the rationals in $[0, 1]$ like so: $$0, 1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{5}{6} \ldots$$ Essentially, I'm listing the points in the graph of the ruler function in lexicographic order. It's not a sequence with a "nice" formula, but it is a sequence.

You can make it a tad nicer if you don't care about injectivity:

$$0, 1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \ldots,$$

or indeed just look at dyadic rationals:

$$1, \frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}, \ldots.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.