What are bits in the context of channel capacity units?
in Information Theory, the amount of information in a message is proportional to the logarithm of the probability of occurrence of the message:

$$ I(m) = -C \log\left( P(m) \right) $$

the only difference between different bases of logarithms is a scaling constant.  if $ C = \frac{1}{\log(2)} $, that is

$$ \begin{align}
I(m) &amp; = -\frac{1}{\log(2)} \log\left( P(m) \right) \\
     &amp;= - \log_2\left( P(m) \right) \\
 \end{align} $$

then we say that the $I(m)$ is in units of bits.  the reason why is if the message $m$ had only two possibilities and both had equal likelihood, then $P(m) = \frac12 $ and $I(m) = 1$.  there are two messages, two possibilities of equal likelihood, like heads and tails of a coin.  and you need only one bit, 0 or 1, to represent the two possibilities.