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"Uniform distribution on [0,1]" is standard terminology for Lebesgue measure. And if m is Lebesgue measure, then m([0,1])=1 and m({x})=0 for all x. The statement of the problem makes perfect sense and the existence of the uniform distribution on [a,b] is proven in any first-year graduate real analysis course, and some undergraduate ones.
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You're joking, right?
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No. It's standard terminology.
Mathworld:
uniform distribution
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When I said "You're joking, right?" I was questioning his assertion that the existence of the Lebesgue measure implies that the problem makes sense. It doesn't.
I already clarified exactly what I meant when I used the phrase "uniform distribution." We need a probability measure on [0,1] that assigns nonzero equal mass to all the singletons. The Lebesgue measure almost satisfies this, but it assigns zero mass to each singleton.
I'm not a probabilist. I wanted to state that it's impossible to select real numbers uniformly at random. I used the phrase "uniform distribution." If that's not the right phrase it doesn't matter because I clarified exactly what I meant.