Your example with G(f,x) = 0 indicates that you don’t understand the solution. An important point to the solution is that the representative g that is chosen is the same regardless of what x is. Otherwise we could not assert that the guesser is wrong for only finitely many x because g(x) differs from f(x) at only finitely many places. In your example, you would have g(x) be identically zero, which obviously does not match f(x) at all but finitely many places.

Your probability argument doesn’t hold because G(f,x) is not a random variable.

Suppose that f is white noise with a non-zero variance. Each f(x) is random, and f(x) and f(x’) are independent of one another if x’ =/= x. Since G(f,x) is a deterministic mapping, and since it uses information about f everywhere except at x, G(f,x) and f(x) are independent. By independence, the variance of the error VAR[G(f,x) - f(x)] for a given x is at least VAR[f(x)]. It follows that the probability of a correct guess (even over random x) is zero.

The article’s proof seems to be based in the statement “the set {x | g(x) differs from f(x)} has measure 0,” which is why a random selection of x helps. However, I believe it is implicit in the proof is that f is fully known to the guesser — it’s fixed at the beginning of the problem and used to construct a “nice” G(f,x) that yields G(f,x) = f(x) except possibly at a finite number of points.

If the guesser didn’t know f, G(f,x) couldn’t be chosen so nicely. Suppose f is non-zero except at finitely many points. For any given f and x, choose G(f,x) = 0. The function g with g=f except at x where we define g(x) = 0 certainly belongs to the same class as f, so we can make it the representative function for that class. However, fix f and define g(x) = G(f,x). Now, g=0, which means that g and f differ at infinitely many points. The probability of guessing f(x) is now zero.

Another way to linguistically slip past this subtlety might have been to say: “The axiom of choice implies that you have a strategy such that, whatever Bob’s distribution over the space of functions, the expected value over that distribution of the probability of you winning the game is 1.”

I think it’s more obvious if you assume at the outset that Bob knows all the equiv class representatives. (giving him more info will only strengthen the proof). So Bob picks places to change the representative, hoping you will stumble on one of those, but of course you don’t.

On the other hand, it really seems magical if you explain it this way:
After Bob picks f, you pick x, and Bob gives you the table, Step 4 is to “pick an equivalence representative g for the table and guess f(x)=g(x)”
all at once.

It could even be made into a joke!