Measurable Spaces – Problem (48/365)

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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