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

is continuous on the right but is not a distribution function in .

Take the first function. To show that it is continuous on the right, let and let . We need to show that there exists such that for all and within a distance of the following holds: . If , then let be the distance to the nearest point where . We see that in this case, picking a point within of will take on a value of and the difference will be less that . If , then and we can easily pick such that , meaning it will also take on a value of and satisfy . 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 because if and the difference function evaluates to .

The second function is not a distribution function because . But it is continuous on the right because if we pick a point the function is constant on the interval and it is open on the right meaning we can always find a delta on the right to satify any .