Consider the Stern-Brocot scheme for generating the positive rationals (with 0 and infinity thrown in): 0/1 < 1/0 0/1 < 1/1 < 1/0 0/1 < 1/2 < 1/1 < 2/1 < 1/0 0/1 < 1/3 < 1/2 < 2/3 < 1/1 < 3/2 < 2/1 < 3/1 < 1/0 etc. (To obtain the n+1st row, we take the nth row and insert the mediant (a+c)/(b+d) of any two adjacent entries a/b and c/d.) Based on this scheme, let's say that 1/2 of the positive rationals are beween 0 and 1, 1/2 of the positive rationals are beween 1 and infinity, 1/4 of the positive rationals are beween 0 and 1/2, 1/4 of the positive rationals are beween 1/2 and 1, 1/4 of the positive rationals are beween 1 and 2, 1/4 of the positive rationals are beween 2 and infinity, 1/8 of the positive rationals are beween 0 and 1/3, etc. If you're suspicious of non-countably-additive measures, you can put a probability measure mu on the positive REALS that assigns measure 1/2 to the intervals [0,1] and [1,infinity], measure 1/4 to the intervals [0,1/2], [1/2,1], [1,2], and [2,infinity], etc. (In fact, the restriction of mu to [0,1] is nothing other than the pullback of Lebesgue measure on [0,1] via Minkowski's question mark function, if you want to be a sophisticate. Alternatively, if you're a fan of the Stern enumeration of the positive rationals (rediscovered by Calkin and Wilf), you can just say, as Reznick does, that the mu-measure of an interval [a,b] is the limit, as n goes to infinity, of the proportion of the first n terms of the Stern sequence that fall in [a,b].) This measure mu has two really nice properties: mu(1/S) = mu(S) and mu(S+1) = mu(S)/2. The first of these you can read off from the 180-degree rotational symmetry of each row of the Stern-Brocot scheme (exchange left with right and numerators with denominators). To see why the second is true, it suffices to note that the last half of each row (leaving aside the final term) is gotten from the previous row (again leaving aside the final term) by adding 1 to each fraction.

From this, it's easy to show that the average positive rational is 3/2!

One way is to note that the center of mu-mass of each interval [n,n+1] is at n+1/2. So the center of mu-mass of [0,infinity) is (1/2)(1/2) + (1/4)(3/2) + (1/8)(5/2) + (1/16)(7/2) + ... and you'll see instantly that this is 3/2 if you're von Neuman. Alternatively, note that the center of mass of [1,infinity) is 1 to the right of the center of mass of [0,infinity) (this follows from the formula mu(S+1) = mu(S)/2). So if we split [0,infinity) into two equal-mass sub-pieces [0,1] and [1,infinity), we get (*) x = (1/2)(1/2) + (1/2)(x+1), where x is the center of mass of [0,infinity), 1/2 is the center of mass of [0,1], and x+1 is the center of mass of[1,infinity).

From (*) we get the solution x = 3/2.

Can anyone see how to compute the center of mass of [0,1/2]? For that matter, what is mean-squared positive rational? Jim