Measure of Irrationals with Even First Digit

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I just finished a course where we discussed concepts such as Lebesgue integration and Lebesgue measure of sets. Today, I was telling my brother about how the irrationals on the interval [0,1] have measure 1, which is sort of counter-intuitive.

Anyway, he proposed the following question. Let A be the set of irrationals on the interval [0,1] whose first digit in their decimal expansion is even. What is the measure of A? Intuitively, I feel like it should have measure 1/2, since it should capture "half" of the irrationals on [0,1]. However, I can't think of any way to cover these irrationals with open intervals of any total length less than 1.

So if A does have measure 1/2, how can we prove that? If A has measure 1, then how do we reconcile this with the fact that the measure of the irrationals on [0,1] is 1?

Thanks!
 
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Can you find any intervals in [0,1] that don't contain any such irrational numbers?
 
Hahaha, of course. Man, it is clear that I haven't slept for a while. Thanks, Hurkyl.

Edit: Well, since my first question was pretty stupid, I have a new one to ask. Is it possible to divide up the irrationals from [0,1] into two sets A and B such that each set contains "half" of the irrationals, but for any open interval O contained in [0,1], m(OnA) and m(OnB) are both non-zero?
 
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Start by taking two dense null sets (the rationals and an irrational translate) then just split the rest of the reals up between the two sets using whatever measure 1/2 breakdown you want (like the one described here)
 
He wanted non-null, not non-empty.
 
Hurkyl said:
Can you find any intervals in [0,1] that don't contain any such irrational numbers?

Surely I am missing some simple points, but what about say [.11, .12]? Every real within this interval must start with 1 and so does not contain any such number?
 
frillth said:
however, i can't think of any way to cover these irrationals with open intervals of any total length less than 1.

(0,0.1)u(0.2,0.3)u(0.4,0.5)u(0.6,0.7)u(0.8,0.9)?
 
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