Why it is not $O(m)$ but $O(\log m)$? I am reading the lecture notes and have a question. I am trying to understand the beginning of Section 3 on page 2. 

Problem: Given an input stream $\sigma$, compute (or approximate) its length $m$.

Naive solution: $O(\log m)$ bits, exact solution.

I don't understand why it is not $O(m)$ bits but $O(\log m)$ bits. Any help would be greatly appreciated.
As a definition, the input size of a problem is the minimum number of bits $(\{0,1\})$ needed to encode the input of the problem. Since it is given that input length $|\sigma|=m$ , you should have $Ceil ( {\log_2 m}) $ bits to encode $\sigma$.