Let us closely examine what your construction actually proves. You want to argue that if you build f by choosing values at different points independently at random, then f(c) is independent of f|

_{R\{c}}. Let's see what happens when we try to make that argument rigorous.

You

can build f by choosing values at different points independently at random: you can view the space

R^{R} of functions from

R to

R as a product space, and a probability measure on

R will induce

a probability measure γ on the product. You can pick a function f at random from this measure space, which does correspond to picking the various values independently at random. For example, it's possible to show that given finitely many points, the distributions of the values of f at those finitely many points are independent, and each matches the original distribution on

R.

We want to show that this choice of f defeats Qaanol's strategy. Showing that we beat Qaanol's strategy with probability 1 is equivalent to showing that

[math]\int_{\bf R^R \times R} \delta(S(f,c),f(c)) \, d(\gamma \times \mu)(f,c)=0,[/math] where μ is the probability measure from which Qaanol picks c, γ is the probability measure on

R^{R}, δ is Kronecker's delta function, and S(f,c) encodes Qaanol's guess for f(c) given f|R\c. The idea would be to show that i) by Fubini's theorem, this integral is equal to the integral [math]\int_{\bf R} \int_{\bf R^R} \delta(S(f,c),f(c)) \, d\gamma(f) \, d\mu(c),[/math] and ii) since the values of f are chosen independently at random, for each c, [math]\int_{\bf R^R} \delta(S(f,c),f(c)) \, d\gamma(f)=0[/math] since the product measure γ factors into the product of a probability measure on

R^{R\{c}} determining f|

_{R\{c}} and a probability measure on

R^{{c}} determining f(c) (which again uses Fubini's theorem). Assuming i) and ii), this tells us that the Qaanol's guess is almost-surely wrong. But as the strategy determined by Qaanol produces a nonmeasurable function S, both steps fail, as Fubini's theorem applies only to measurable functions. (The proof does, however, show that no measurable strategy can succeed.)

Mike-l was completely correct when he pointed out the connection with nonmeasurable sets.