The sets with this property (that you can tell whether your number is in or out after only finitely many dice rolls) are the open sets, not the measurable sets.
That doesn't seem right to me. Take as my target the open set (0, pi-3). If I keep rolling zeros I'll never be able to stop. (Edit: I know that the probability of rolling all zeros approaches 0 as the number of die rolls approaches infinity, but I assume that a demon can take over the die and start feeding me all zeros, or the digits of pi-3 or whatever. As I think about this more I'm thinking maybe what you said works if there is no demon. Edit 2: Or not. If there's no demon and my first digit is 0 then I can stop, but that's only because 0 is expressible as an integer divided by a power of ten. If there's no demon and I roll the first few digits of pi-3, I know that I'll eventually go over or under pi-3, but I don't know which, and it doesn't matter whether pi-3 itself is in my target set.)
Every die roll tells me that the random number I'm generating lies in the closed interval [x, x+1/10^n], where x is the decimal expansion I've generated so far and n is how many digits I've generated. If at some point I start rolling all 0s or all 9s I'll be rolling forever if the number I'm generating is a limit point of the target set, even if it's not in the target set.
I should have been more accurate and said "If the random number that you'll eventually get does in fact lie in the set, then you'll find out about this fact after a finite number of rolls."
This really does define open sets, since for any point in an open set there's an open ball of radius epsilon about it which is in the set, and then the interval [x, x+1/10^n] has to be in that ball once 1/10^n < epsilon/2.
EDIT: (and the converse also holds, I think, but it requires some painfully careful thinking because of the non-uniqueness of decimal expa...
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