Measurable Spaces – Problem (48/365)

Posted on November 6, 2016 by Naren Sundar

Another question on distribution functions. Show that each of the functions

\displaystyle G(x,y) = 1 \text{ if } x+y \ge 0 \text{ and } 0 \text{ otherwise} \\ G(x,y) = [x+y] \text{, the integral part of } x+y

is continuous on the right but is not a distribution function in \mathbb{R}^2.

Take the first function. To show that it is continuous on the right, let \epsilon > 0 and let (x,y) \in \mathbb{R}^2. We need to show that there exists \delta > 0 such that for all (a,b) > (x,y) and within a distance of \delta the following holds: | G(x,y) - G(a,b) | < \epsilon. If G(x,y) = 0, then let \delta be the distance to the nearest point (p,q) where p+q = 0. We see that in this case, picking a point (a,b) > (x,y) within \delta of (x,y) will take on a value of 0 and the difference will be less that \epsilon. If G(x,y) = 1, then x \ge y and we can easily pick (a,b) > (x,y) such that a > b, meaning it will also take on a value of 1 and satisfy \epsilon. Thus, the first function is continuous on the right.

However, it is not a distribution function because it does not satisfy the requirement described in the last post that \Delta_{a_1b_1}\Delta_{a_2b_2} \ge 0 because if (a_1,a_2) = (1,2) and (b_1,b_2) = (-4,3) the difference function evaluates to -1.

The second function G(x,y) = [x+y] is not a distribution function because G(\infty, \infty) \ne 1. But it is continuous on the right because if we pick a point (x,y) the function is constant on the interval [[x+y],[x+y+1]) and it is open on the right meaning we can always find a delta on the right to satify any \epsilon > 0.

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