TITLE: What does "the average continuous function is nowhere monotonic" mean?
QUESTION [7 upvotes]: I plan on asking my professor what he meant by "average continuous function," but as it is possible that this is a concept as vague as the statement, I was hoping to get some interesting answers/interpretations from stack exchange first.
How do you think of the average of some infinite group of things? 
Or does this just mean that the real line is so dense/big that it is somehow likely that a function would bounce around everywhere except on some countable number of points?
I'm sorry this is vague, I will be sure to post his response if I get a good one.
I would also appreciate any resources or reading; googling around hasn't been fruitful.

REPLY [6 votes]: Here is one possible interpretation of what he meant: every continuous, real-valued function on an interval can be approximated as well as you like by continuous, nowhere monotonic functions. In other words, these functions are dense.
In more detail, consider the space $C([0,1])$ (for example) of continuous functions $[0,1]\to\mathbb R$. To talk about the nowhere monotonic functions being dense, the space $C([0,1])$ needs to have some sort of topology, or better yet, a metric. Given $f,g\in C([0,1])$, say that the distance $d(f,g)$ between $f$ and $g$ is
$$
d(f,g) = \sup_{t\in[0,1]} \big|f(t) - g(t)\big|.
$$
(This is a metric!)

Theorem: Under this metric, the space of continuous, nowhere monotonic functions is a dense subspace of $C([0,1])$.

In other words, I can approximate continuous functions on the interval by nowhere monotonic functions:
if you give me some random continuous function $f:[0,1]\to\mathbb R$, and some tolerance $\varepsilon>0$, I can find a nowhere monotonic function $g:[0,1]\to\mathbb R$ such that $|f(t)-g(t)|<\varepsilon$ for all $t\in[0,1]$.
This theorem has an easy proof using the Baire Category Theorem. Your professor can probably give it, or one of us at MSE can if you would like.
Denseness is only one notion of a behavior being "average", and it's not always the best one. For instance, the rationals are a dense subset of the reals even though the "average" real number is irrational. Several other answers give alternative notions of average.