Consider the following example for the interval X = (0, 1) (which is homeomorphic to R). Suppose we wanted to assign measures to all of its subsets, and do so in accordance with the ordinary desiderata of sigma-additivity and m(X) = 1.
Now partition the interval into an uncountable family of countable sets X_i such that two numbers live in the same subset iff they differ by a rational number. (Make sure you fully understand this construction before continuing!)
What measure should we assign to any such X_i? We can quickly see that the X_i are all of equal cardinality (that of Aleph-null) and even have natural maps to each other (given by adding irrationals mod 1).
We can’t assign them measure 0 - by sigma-additivity that gives us m(X) = 0. We can’t assign them positive measure—again by sigma-additivity that gives us m(X) >> 1.
Thus we cannot assign such subsets any measure, so we must have been wrong from the start that 2^X was a reasonable sigma-algebra to pick as the foundation of our measure in X.
I fail to see why the family of sets Xi is countable. if Xi is of cardinality ℵ0, which I totally agree about, then how can a union of a countable family of them which is basically ℵ0×ℵ0 be equal (0,1)?
Though now that I think about it, if the difference is some irrational number then this seems to work, as any set Xi would contain exactly one unique rational number. Now they each have the cardinality of R, and the family has the cardinality of Q. And then it all seems to work. Does that seem right?
Consider the following example for the interval X = (0, 1) (which is homeomorphic to R). Suppose we wanted to assign measures to all of its subsets, and do so in accordance with the ordinary desiderata of sigma-additivity and m(X) = 1.
Now partition the interval into an uncountable family of countable sets X_i such that two numbers live in the same subset iff they differ by a rational number. (Make sure you fully understand this construction before continuing!)
What measure should we assign to any such X_i? We can quickly see that the X_i are all of equal cardinality (that of Aleph-null) and even have natural maps to each other (given by adding irrationals mod 1).
We can’t assign them measure 0 - by sigma-additivity that gives us m(X) = 0. We can’t assign them positive measure—again by sigma-additivity that gives us m(X) >> 1.
Thus we cannot assign such subsets any measure, so we must have been wrong from the start that 2^X was a reasonable sigma-algebra to pick as the foundation of our measure in X.
I fail to see why the family of sets Xi is countable. if Xi is of cardinality ℵ0, which I totally agree about, then how can a union of a countable family of them which is basically ℵ0×ℵ0 be equal (0,1)?
The number of such sets is specifically uncountable. Each set is of itself countable. Apologies, I’ll fix the OP.
Though now that I think about it, if the difference is some irrational number then this seems to work, as any set Xi would contain exactly one unique rational number. Now they each have the cardinality of R, and the family has the cardinality of Q. And then it all seems to work.
Does that seem right?
That sounds like it also works. I’ve seen the proof both ways and I think I was mixing them together in my head.
That sounds like “you can’t pick a uniformly random natural number”, and yet that’s how I’d motivate limits.