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Show that the set of irrationals in $[0,1]$ does not have measure zero in $\mathbb{R}$

By definition of measure zero, we must show that there exists $\epsilon>0$ such that the set $A$ of irrationals in $[0,1]$ cannot be covered by a countable collection of intervals $[a_1,b_1],[a_2,b_2],\ldots$ with total length less than $\epsilon$.

So I take $\epsilon=1/2$, and suppose for contradiction that the countable collection of intervals $[a_1,b_1],[a_2,b_2],\ldots$ covers $A$ and $(b_1-a_1)+(b_2-a_2)+\ldots < 1/2$. How can I continue?

EDIT: Thanks for all the responses. Actually I haven't learned any measure theory, only the definition of measure zero in real analysis. Is it possible to proceed using only that? (By the way, sure, I can show $\mathbb{Q}\cap[0,1]$ has measure zero.)

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    $\begingroup$ Why not use the fact that $m([0,1]) = m(\mathbb{Q}\cap[0,1]) + m(\mathbb{Q}^c\cap[0,1])$? $\endgroup$ Aug 30 '13 at 17:45
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    $\begingroup$ Can you show a countable set (in particular, $\Bbb Q\cap[0,1]$) has measure zero? If so, then the rest follows $\endgroup$
    – Clayton
    Aug 30 '13 at 17:45
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    $\begingroup$ Actually I haven't learned any measure theory, only the definition of measure zero in real analysis. Is it possible to proceed using only that? (By the way, sure, I can show $\mathbb{Q}\cap[0,1]$ has measure zero.) $\endgroup$
    – Mika H.
    Aug 30 '13 at 17:47
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Ok, you have a collection $[a_1, b_1],...,[a_n, b_n]$ that covers $A$ of total diameter less than $\frac{\epsilon}{2}$. You know that the rationals in $[0,1]$ are measure $0$, so you can find $[c_1, d_1], ..., [c_m, d_m]$ covering $\mathbb{Q} \cap [0,1]$ of diameter less than $\frac{\epsilon}{2}$. Then together, these intervals form a cover of $[0,1]$ of total diameter less than $\epsilon$ which is a contradiction.

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