Continuity of the Lebesgue function If $x \in [0,1]$ has ternary expansion $(a_n)$, i.e. $x = 0.a_1a_2..$ with $a_n =0,1$ or $2$, define $N$ as the first index $n$ for which $a_n = 1$, and set $N = \infty$ if none of the $a_n$ are $1$ (when $x \in $ Cantor Set). Now, set $b_n = \frac{a_n}{2}$ for $n < N$ and $b_n = 1$, and let $$F(x) = \sum_{n=1}^{N} \frac{b_n}{2^n}$$
for each $x \in [0,1]$ How does its partial graph look like? $F$ seems continuous but I have no idea how to prove it. Can someone help me with this? Thanks.
 A: The cleanest way to do this that I know is to show that $F$ is the uniform limit of a sequence of continuous functions, and is hence continuous.
Consider the sequence $C_k$ of closed sets which go in to the construction of the Cantor set $C$ (ie. $C_k$ is the union of closed intervals that are remaining at the $k^{th}$ stage). Thus
$$
C = \cap_{k=1}^{\infty} C_k
$$
Now let $D_k = [0,1]\setminus C_k$ which is a union of $2^k - 1$ intervals $\{I_j^k\}$, ordered from left to right. Define $f_k:[0,1]\to \mathbb{R}$ be the continuous function defined such that
$$
f_k(0) = 0, \qquad f_k(1) = 1
$$
$$
f_k(x) = \frac{j}{2^k} \quad \forall x\in I_j^k, 1\leq j \leq 2^k -1
$$
and which is linear on each interval of $C_k$. (Try drawing the pictures for $f_1$ and $f_2$ to get an idea as to what is going on).
Now $\{f_k\}$ is monotone increasing, and
$$
f_{k+1} = f_k \quad \text{on} \quad I_j^k, 1\leq j\leq 2^k -1
$$
Hence,
$$
|f_{k+1} - f_k| < 2^{-k}
$$
Hence, $\sum (f_{k+1} - f_k)$ converges uniformly on $[0,1]$, and so
$$
F = \lim_{k\to \infty} f_k
$$
is also continuous.
Edit: I just found the following picture online, which explains what $f_1$ and $f_2$ look like (The picture has an added $f_0$ function as well):

