On dyadic numbers

A real number $x$ is said to be dyadic rational provided there is an integer $k$ and a non negative integer $n$ for which $\displaystyle x=\frac{k}{2^n}$ . For each $x \in [0, 1]$ and each $n \in \mathbb{N}$ set:

$f_n(x) = \left\{\begin{matrix} 1 &, & x =\dfrac{k}{2^n} , \; k \in \mathbb{N} \\\\ 0& , & \text{otherwise} \end{matrix}\right.$

(i) Prove that the dyadic numbers are dense in $\mathbb{R}$ .

(ii) Let $f:[0, 1] \rightarrow \mathbb{R}$ be the function to which the sequence ${f_{n}}_{n \in N}$ $\{f_n\}_{n \in \mathbb{N}}$ converges pointwise. Prove that $\bigintsss_0^1 f(x) \, {\rm d}x$ does not exist.

(iii) Show that the convergence $f_n \rightarrow f$ is not uniform.

Solution

(i) Let $a, b \in \mathbb{R}$ such that $a<b$ . We want to show that there exists a dyadic rational in the open interval $(a, b)$ . By the Archimeadean property we can choose a positive integer such that $2^n (b-a) >1$ so that we know that there is an integer $k$ such as $2^n a <k <2^n b$ hence $\displaystyle a<\frac{k}{2^n}< b$ . Now we can easily conclude that the dyadic numbers are dense in $\mathbb{R}$ .

(ii) If $\displaystyle x=\frac{k}{2^n} \in [0, 1]$ then $\displaystyle x=\frac{k 2^r}{2^{n+r}}$ forall $r>0$ . This says that $f_m(x) =1$ forall $m>n$ . If $x$ is not dyadic rational then $f_n(x)=0$ forall $n \in \mathbb{N}$ . Hence the limit of the function is:

$f(x)= \left\{\begin{matrix} 1 &, & x \; \text{is a non zero dyadic rational} \\\\ 0& , & \text{otherwise} \end{matrix}\right.$

Let us now consider a partition $\mathcal{P}=\{x_0, x_1, \dots, x_n \}$ of $[0, 1]$ . By the density of the dyadic numbers $\sup \left \{ f(x): x \in [x_{i-1}, x_i] \right \}=1$ for each subinterval of the partition. This says that the upper integral of $f$ is equal to $1$ . Similarly, since the irrationals are dense in $[0, 1]$ it follows that the lower integral of $f$ is $0$ . Thus the requested integral does not exist.