Is there any sense in which the monte-carlo approximation converges if and only if the integral exists? I'm thinking of the case where f is the indicator function of the irrationals, and the domain is [0, 1]. Then I think the Monte-Carlo approximation would converge to 1, since a uniformly chosen number in [0, 1] maps to 1 under f with probability 1, but the (Riemann) integral is undefined.
Does this issue get fixed with Lebesgue integration?
Is there any sense in which the monte-carlo approximation converges if and only if the integral exists? I'm thinking of the case where f is the indicator function of the irrationals, and the domain is [0, 1]. Then I think the Monte-Carlo approximation would converge to 1, since a uniformly chosen number in [0, 1] maps to 1 under f with probability 1, but the (Riemann) integral is undefined.
Does this issue get fixed with Lebesgue integration?