real analysis - Prove that f(x) is not continuous and differentiable at any point. Find Riemann sums.

I would like some feedback and help on the following homework problem for Real Analysis.

Consider the function f : $\mathbb{R}$ → $\mathbb{R}$ given by
$$
f(x) =
\begin{cases}

\lfloor x\rfloor, & x\ \text{is rational},\\
\lfloor x\rfloor+3, & x\ \text{is irrational}
\end{cases}
$$

where $\lfloor x\rfloor$ is the greatest integer less than or equal to x.

---

1. Prove that $f(x)$ is not continuous at any point.


2. Prove that $f$ is not differentiable at any point.



3. Consider an arbitrary partition P = $\{x_0, x_1, x_2, \cdots, x_n$}$ of $[1, 2]$.

   
   
   

   a) Find $U(f, P)$ and $L(f, P)$.

   
   
   

   b) Find $U(f)$ and $L(f)$.

   
   
   

   c) Prove that f is not integrable on $[1, 2]$.

---

For #1, I let $c\in\mathbb{R}$ and found the limit of the sequence of rationals $s_n$ converging to c and that equals to $\lfloor c\rfloor$. Then the limit of the sequence of irrationals $t_n$ converging to c and got $\lfloor c\rfloor+3$. Now for f to be continuous at c, the two limits must be the same, but clearly $\lfloor x\rfloor \neq \lfloor x\rfloor + 3$. Therefore, $f$ is not continuous at any point.

For #2 & 3 I'm stuck on.

• How do I show the limits do not exist?


• Is it correct if I say "by definition, f is not continuous at any point so it's not differentiable at any point"?



• How do I find the upper & lower Riemann sums for #3 if I don't have values to plug in? What would $\Delta x_i$ be?

Any help is appreciated. Thank you so much!

Answer

For 2, you want to show that
\begin{align*}
\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}
\end{align*}
doesn't exist for any $x\in\mathbb{R}$.

The easiest way of doing this is by considering the sequence $(a_n) = \sqrt{2}/n$. Then this is a sequence which is irrational at every point and converges to zero. Thus if $\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$ exists, it must be equal to:
\begin{align*}
\lim_{n\rightarrow\infty}\frac{f(x+a_n)-f(x)}{a_n}
\end{align*}
For any rational $x$, this limit does not exist, as it tends to $3/0$.

Similarly for irrational $x$, you must must use a sequence of numbers converging to zero so that $x+a_n$ is always rational. This is a bit more abstract.

Choose $x$ irrational. Consider a monotone decreasing sequence $(x_n)$ tending to zero but equal to zero for any $n$ (e.g. $1/n$). Then $x < x+x_n$ for each $n$. Hence for each $n$ you can find a rational number $y_n$ so that $x < y_n < x+x_n$. Now the sequence $(b_n) = (y_n-x)$ is also converging to zero, and has the property that $f(x+b_n)$ is rational, while $f(x)$ is irrational, for your given $x$.

Now you can use a similar argument as the first case to show that the limit as $h$ tends to zero of $\frac{f(x+h)-f(x)}{h}$ does not exist.

For 3: First let's think about the definitions.
$U(f, P)$ is the upper sum for this partition, so it's defined to be the sum of all $\sup\{f(x):x\in[x_i, x_{i+1}]\}(x_{i+1}-x_i)$. This is essentially "giving you the volume of the box" above the graph. Because there is an irrational number between any two distinct real numbers, you know that $\sup\{f(x):x\in[x_i, x_{i+1}]\}$ must be $|x|+3 = 4$, regardless of $x_i$ or $x_{i+1}$. Then you get
\begin{align*}
\sum_{i=0}^{n-1}4(x_{i+1}-x_i) = 4(x_n-x_0) = 4
\end{align*}
Similarly you can show that $L(f, P) = 1$.

For part (b), can you see that the partition didn't have any effect on the result for (a)? This means you can prove that as the partition gets arbitrarily fine, your upper and lower sums will still be 4 and 1.

Finally this results tells you that $U(f) \neq L(f)$, so it's not Riemann integrable.