$f(x)=\lim f_n(x).$

Question

A question from Introduction to Analysis by Arthur Mattuck:

Let $f_n(x)=\begin{cases}1, & \text{if $x$ can be written $i/2^n$ for some integer $i$;} \\0, & \text{otherwise.} \end{cases}$

Answer the following, with proofs.

(a) Let $f(x)=\lim f_n(x).$ What is $f(x)$ explicitly ?

(b) Is the convergence $f_n \to f$ uniform on $I=[0,1]$?

(c) Is $f_n$ Riemann-integrable? Is $f$ ?

I am not sure what $f$ is.

Answer 1

Well, note that if $f_N(x_0)=1$ for some $N$, then also $f_n(x_0)=1$ for all $n>N$; so to get $f(x_0)=1$, we need some $N$ with $f_N(x_0)=1$. This essentially means,

If $f(x)=1$, then there exists integers $i$ and $N$ such that $x=i/2^N$.

Since $f$ only takes values $0$ and $1$, we get

$$f(x)=\begin{cases}1, & \text{if $x$ can be written $i/2^n$ for some integers $i$ and $n$} \\0, & \text{otherwise.} \end{cases}$$

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Do you see how this translates to uniform convergence? Do you think this can be integrated? What do you think the graph looks like?