Finding Function's Extension and Its Unique Existence. Let $$A= \left\{\frac j{2^n}\in [0,1] \mid n = 1,2,3,\ldots,\;j=0,1,2,\ldots,2^n\right\} $$ and let $$ f:A\rightarrow R    $$ satisfy the following condition:
There is a sequence $ \epsilon_n \gt 0 $ with $\sum_{n=1}^\infty \epsilon_n \lt \infty $ and $$\left|f\left(\frac{j-1}{2^n}\right)-f\left(\frac j{2^n}\right)\right| \lt \epsilon_n $$ for all $ n\gt0, j=1,2,\ldots,2^n $.
Prove that $f$ has a unique extension to a continuous function from $[0,1]$ to $R$.

I think it is maybe related to contraction mapping.
But, I could not apply that thery into this problem.
Is there anybody who can prove this? 
p.s. The Formatting is helped by @Math1000. I appreciate that.
 A: An alternate answer that uses some of Alex Ravsky's method, but sticks to analysis methods, and in my opinion provides a more constructive demonstration:
For $x \in [0, 1]$, let $$x = \sum_{k = 1}^{\infty} \omega_{k}(x) 2^{-k},\; \omega_{k} \in \{0, 1 \}$$ i.e. suppose $(\omega_{k}(x))_{k \in \mathbb{N}}$ is the binary expansion of $x$, and define the extension of $f$ by $$f(x) = \lim_{n \to \infty} f\left(\sum_{k = 1}^{n} \omega_{k}(x) 2^{-k}\right).$$ We claim first that the sequence converges by showing it to be Cauchy for a given $x \in [0, 1]$, which we now fix. Let $\epsilon > 0$, and pick $N \in \mathbb{N}$ such that if $n \geq N$, then $\sum_{k = n}^{\infty} \epsilon_{n} < \epsilon$. Pick $m, n \geq N$. Then as Ravsky points out,
\begin{align*}
\left|f\left(\sum_{k = 1}^{n}\omega_{k}(x) 2^{-k}\right) - f\left(\sum_{k = 1}^{m}\omega_{k}(x) 2^{-k}\right)\right| & < \epsilon
\end{align*}
Boom. Cauchy, and thus convergent, so our extension is defined. Now, let $|y - x| < 2^{-(N + 2)}$. Having chosen $x$ and $y$ close enough, we have that $\omega_{k}(x) = \omega_{k}(y)$ for all $k \leq N$. Thus $$|f(x) - f(y)| \leq \sum_{k = N + 1}^{\infty} \epsilon_{k} < \epsilon,$$ completing the proof of continuity.
