MHB Advanced Calculus - Continuous Functions

[bradyrsmith31]

I'm really stumped on how to do these proofs…

I would really appreciate any help or insight!

 

[Euge]

Hi bradyrsmith31,

For part (a), take any $x \in [0,1] \cap \Bbb Q$ and let $x_n := \frac{x}{2^{1/2^n}}$. Then $x_n \in [0,1]\setminus \Bbb Q$ such that $x_n \to x$. Since $f(x_n) = 0$ for all $n$ and $f(x) > 0$, $f(x_n)$ does not converge to $f(x)$. Therefore, $f$ is discontinuous at $x$.

For part (b), let $\varepsilon > 0$ and $x_0 \in [0,1] \setminus \Bbb Q$. Since $\lim_{n \to \infty} a_n = 0$, there exists a positive integer $N$ such that for all $n \ge N$, $a_n < \varepsilon$. Partition $[0,1]$ into $N$ subintervals $J_1,\ldots, J_N$ of length $\frac{1}{N}$. Let $k$ be the index such that $x_0 \in J_k$. Set $\delta$ equal to the distance from $x_0$ to the nearer endpoint of $J_k$. If $|x - x_0| < \delta$, then $x$ lies in the interior of $J_k$. If $x$ is rational, this implies $d(x) \ge N$. Consequently, $$|f(x) - f(x_0)| = |f(x) - 0| = f(x) = a_{d(x)} < \varepsilon.$$ If $x$ is irrational, then $|f(x) - f(x_0)| = |0 - 0| = 0 < \varepsilon$. Since $\varepsilon$ and $x_0$ are arbitrary, $f$ is continuous at every point of $[0,1]\setminus \Bbb Q$.

For part (c), note that $f$ is bounded and continuous outside a null set.

For part (d), consider any partition $P : 0 = x_0 < x_1 < \cdots < x_n = 1$ of $[0,1]$. The lower Riemann sum $L(f,P)$ is zero because each interval $[x_{i-1}, x_i]$ contains an irrational point $q_i$, and $f(q_i) = 0 \le \inf_{t \in [x_{i-1},x_i]} f(t)$. Consequently, the lower Riemann integral of $f$ is zero.

The upper Riemann integral of $f$ is also zero. For let $\varepsilon > 0$. Let $N$ be a positive integer such that $a_n < \frac{\varepsilon}{2}$ for all $n \ge N$. Partition $[0,1]$ into the intervals $I_k :=[\frac{k-1}{N}, \frac{k}{N}]$, for $k = 1, 2,\ldots, N$. For each $k$, let $M_k := \sup_{t \in I_k} f(t)$. There exist $r_1,\ldots, r_n$ such that $t_k \in I_k$ such that $M_k < f(r_k) + \frac{\varepsilon}{2}$ for all $k$. Given $k$, if $r_k$ is rational, $f(r_k) = 0$; if $r_k$ is irrational, $d(r_k) \ge N$ and thus $f(r_k) = a_{d(r_k)} < \frac{\varepsilon}{2}$. Thus $M_k < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ for each $k$. The upper Riemann sum $U(f,P) = \frac{1}{n} \sum_{k = 1}^n M_k $ is therefore less than $\varepsilon$. Since $\varepsilon$ is arbitrary, this shows that the upper Riemann sum of $f$ is zero.

 

[johng1]

I don't believe Euge's proof is correct. He seems to be saying that if q is a rational between k/N and k/(N+1), d(q) is greater than or equal to N. Of course, this is false: 500/1001<1/2<501/1001. In particular, I don't believe his proof of b) in case $x_0$ is "very close" to 1/2.

Here is my proof of the problem. The idea of the proof of c) and d) is fairly simple, but the details are a little messy.

 

[Euge]

Hi johng,

You're right about my proof of (b) (and part of (d)). Even in the proof of (d), I mixed up the words "rational" and "irrational" near the end. I meant to fix these issues later tonight, but you beat me to it! In any case, thanks for the critique.