> almost all
> A shorthand term for "all except for a set of measure zero",
> when there is a measure to speak of.
In this case there is no measure to speak of, so it is normal to talk about taking the ration in a finite initial segment and then letting the size go to infinity and seeing what happens to the ratio.And a set of measure zero does not have to be finite, either. Almost all numbers are transcendental in the sense given above, and yet there are infinitely many non-transcendental numbers.
So I think you are mistaken, although I would be interested to hear a more complete statement of what you mean, in case I have misunderstood you.
Put another way, if we stick to the definition of "almost all" as elements not having the property forming measure zero set, then you do not need the video to prove the title: all integers are already measure zero. So, for subsets of integers, "almost all" means something else in jargon.
I could not watch the video, but I found (I believe) equivalent text. And "almost all" means that as we extend the reach the share of numbers not containing 3 goes to zero. It is mildly entertaining (albeit a bit too obvious) result, but my point was that usage of "almost all" is confusing because _I_ am used to a very specific meaning of it.
leastpow2 x = head $ dropWhile (\k -> 2*k <= x) $ map (2^) [0..]
p k = k < 3 * (leastpow2 k `div` 2)
limitless = [k | k <- [0..], k == 0 || p k]
which oscillates between attaining 1/2 and then 2/3 for its partial densities.Indeed it is normal to talk about it, but this condition is usually called having "density 1" (https://en.wikipedia.org/wiki/Natural_density). I think that "almost all" is not terribly common in this connection, except possibly in a community where this is the only kind of nearly-true statement that one would want to discuss.