A question about the Cantor Lebesgue function I am reading some lecture notes on the Cantor Function. Here is the construction that notes used. The notes used the Cantor set to construct a function. Here the Cantor set is defined to be $\{O_n \}$ which consists of deleted middle thirds. i.e.
$$ O_1 = (1/3, 2/3)$$
$$O_2 = (1/9, 2/9) ∪ (1/3, 2/3) ∪ (7/9, 8/9)$$
$$O_3 = (1/27, 2/27) ∪ (1/9, 2/9) ∪ (7/27, 8/27) ∪ · · · ∪ (25/27, 26/27)$$ and so on...
Next, they defined a sequence of functions $\{ \varphi_n\}$ as follows. Each function $\varphi_n$ will satisfy $\varphi_n(0) = 0$ and
$\varphi_n(1) = 1.$ For $n = 1,$ they define $\varphi_1(x) = 1/2$ on $O_1$, and extend it linearly to [0, 1]. For $n = 2,$ they defined
$\varphi_2$ on the components of $O_2$ by $$\varphi_2(x) =
    \begin{cases}
      1/4, & \text{if}\ x \in (1/9,2/9) \\
      2/4, & \text{if}\ x  \in (1/3, 2/3) \\
3/4, & \text{if}\ x  \in (7/9, 8/9) \\
    \end{cases}$$
and extend it linearly to $[0, 1].$ Then, nn the components of $O_n$,
$\varphi_n$ takes values $1/2^n, 2/2^n, . . . ,(2^n − 1)/2^n,$ and extend it linearly to $[0, 1].$ Thus, a sequence of continuous functions $\{ \varphi_n \}$ is formed.
My question here is that, the notes stated "It is obvious that  $| \varphi_{n+1}(x) − \varphi_n(x)| < 2^{−n}$ for every $x \in [0, 1].$ " Can anyone explain to me why this is so obvious? I do not see it at all.
 A: I think it is by definition. The problem here is that the definition is quite complicated and not a simple formula.
Let's look at a point $x \in [0,1]$ and fix $n \in \mathbb{N}$ and distinguish some cases:
If x is in $O_n$ then $$\varphi_n(x) =\varphi_{n+1}(x) = \varphi(x)$$ for $\varphi(x)$ the final function.
Let us now assume $$\varphi_n(x) \leq \varphi_{n+1}(x).$$
We define $y$ as the biggest point in $\overline{O_n}$ with $y\leq x$ and $z$ as the smallest point in $\overline{O_{n+1}}$ with $z \geq x$. Then since $\varphi_n, \varphi_{n+1}$ are monotonously increasing we have
$$
|\varphi_{n+1}(x) - \varphi_{n}(x)| \\
= \varphi_{n+1}(x) - \varphi_{n}(x) \\
\leq \varphi_{n+1}(x) - \varphi_{n}(y) \\
\leq \varphi_{n+1}(z) - \varphi_{n}(y) \\
= 2^{-n-1} < 2^{-n}
$$
by construction of $\varphi_{n+1}$.
If on the other  hand $\varphi_{n+1}(x) < \varphi_n(x)$ we have to take the smallest point in $\overline{O_n}$, the biggest point $\overline{O_{n+1}}$ and so on, but the argument stays the same.
Intuitively this is true because the total "jumps" from n to n+1 are smaller than $2^{-n}$ and the linear part makes the difference even smaller. For $\varphi_n(x) \leq \varphi_{n+1}(x)$ we can even assume that $\varphi_n$ instead of being linear in the gaps of $O_n$ stays constant and only jumps at the very end of the gap, while $\varphi_{n+1}$ does the opposite and jumps at the beginning of the gaps of $O_{n+1}$. Hopefully there are no mistakes in it
