The space you're talking about, the space of all one-way infinite binary sequences, is known as Cantor Space (especially within logic). I think of it as the set of all paths through the full infinite binary branching tree.
As Qiaochu said, the 'fair coin measure' can be thought of as a product measure. But I prefer to just recall that by Caratheodory's extension theorem, it suffices to define any set function $\mu$ on the algebra of basic clopens (that is, sets of the form $[\sigma]:=\{A \in \{0,1\}^{\mathbb{N}} : A \text{ extends } \sigma \},$ where $\sigma$ is some finite binary string) such that $\mu[\sigma] = \mu[\sigma 1] + \mu[\sigma 0]$ and $\mu[\varnothing] = 1$. ("$\varnothing$" denotes the empty finite binary string; all elements of $\{0,1\}^{\mathbb{N}}$ extend it.)
So to get the measure you want, you just define your $\mu$ such that $\mu[\sigma 0] = \mu[\sigma 1] = \mu[\sigma]/2$. Notice that in terms of the tree picture, you're just saying that no matter where you are in the tree, it costs the same amount of measure to go the the left as to the right (so my tree is growing upwards).
Notice that this measure (often called the Lebesgue measure because of what @Michael said in the comments) is just a special case of a Bernoulli measure, which corresponds to a coin with probability $p$ of heads and $1-p$ of tails (or if you prefer $p$ to go left in the tree and $1-p$ to go right).
There are other ways to see that almost every number must have a 'balanced' frequency of 0's and 1's, but none of them are much nicer than using the law of large numbers. For example, one could apply, as @Qiaochu suggested, the Borel-Cantelli lemma. To do so, you want to exploit something that happens infinitely often (because of the $\limsup$). If some $A \in \{0,1\}^{\mathbb{N}}$ does not have a balanced number of 1's and 0's, then frequency of 0's compared to the frequency of 1's is outside of the interval from $[1- 1/n, 1+1/n]$ for some $n$ infinitely often. I haven't checked the whether the resulting sets have measures whose total sum is finite (as is required by the B-C), but something like this should work.
One could also apply Birkhoff's Ergodic Theorem (which is just a generalization of the law of large numbers anyway) or the theory of Algorithmic Randomness to conclude that almost every normal has a balanced frequncy of 0's and 1'. In fact, you can conclude much more, that almost every real number is Normal in every base.