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Let be two functions of x, B(x) and A(x).

Boas ("A primer or real functions", p. 137) says:

If A(x) < B(x), there are rational numbers r and R such that A(x) < r < R < B(x) (OK). Since there are only countable many pairs of rational numbers (OK), the set where B(x) > A(x) is contained in the union of countably many sets of measure zero (why?) ...

I know that a set composed by two numbers has measure zero, and I assume that "there are only countable many pairs of rational numbers", but I fail to see why this implies the next statement ("the set where B(x) > A(x) is contained in the union of countably many sets of measure zero").

Thanks for your help.

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# Union of countable many sets

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