Why do almost all points in the unit interval have Kolmogorov complexity 1?

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I am reading

and I am having a difficult time figuring out an argument they say is obvious.

Define a function $K(x)$ on the unit interval to be $$K(x)=\liminf_{n \to \infty} K(x_n)/n,$$ where $K(x_n)$ is the information constant of $x_n$, the first $n$ bits of $x$ (i.e., the size in bits of the smallest program which will cause a fixed universal Turing machine to produce $x_n$). They make the claim that the set of $x \in \mathbb{R}$ such that $K(x)=1$ has full measure. Why is this? I see that by translation/scaling invariance by rational numbers, it either must have full or zero measure, but I can't make an argument why it must have positive measure. They say it is by a "simple counting argument," which I don't understand.

Note: I re-posted this question to MathOverflow here, where I got a satisfactory answer thanks to Ronnie Pavlov.