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If $A$ is a null set and $B$ is a countable set, both in $\mathbb{R}$ can anyone help me show that $A+B$ is null.

So I'm slightly unsure of where to start here, but how about assuming that $A+B$ is not null. therefore there must exist an interval of real numbers in $A+B$, but that would be uncountable, contradiction. Am I along the right lines here?

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    Thanks for pointing out where I went wrong :)2012-01-21

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Note that A + B = \{ a + b \mid a \in A , b \in B \} = \bigcup_{b \in B} b + A

Then you have:

$ \mu (A + B) \leq \sum_{i = 1}^\infty \mu(b_i + A) = \sum_{i = 1}^\infty \mu (A) = 0$

Where the inequality follows from the countable subadditivity of $\mu$ and you have $\mu(b_i + A) = \mu(A)$ because the Lebesgue measure is translation invariant.

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    @LHS That's very kind of you, thank you. Glad I could help. : )2012-01-21
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Your attempt will not work because it is false that a set of positive measure must contain an interval. If you are familiar with the Cantor set, there is a variation of the construction, giving what are called fat Cantor sets. These sets contain no intervals, but have positive measure. (Or, as Dejan Govc notes, the set of all irrationals).

To prove the statement, note that $A+b$ is null and measurable for each $b\in B$. Since $A+B = \bigcup_{b\in B}(A+b)$ and there are only countably many elements in $B$, it follows that $\mu(A+B) = \mu\left(\bigcup_{b\in B}(A+b)\right) \leq \sum_{b\in B}\mu(A+b).$

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    Oh right I because we're only shifting the null set along it remains null.. you can really tell i've just been trying to teach myself all of this this evening..2012-01-21