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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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Hint: If $A= \{ x_i \}_{i \in N}$, then $A+B= \cup_{i \in N} A+x_i$.. What can you say about a countable union of null sets? – N. S. Jan 21 '12 at 22:29
Hint: write $A + B$ as a countable union of null sets in a natural way. – Zarrax Jan 21 '12 at 22:29
Hi again, N.S ;) – Zarrax Jan 21 '12 at 22:30
Hi again, LOL.. – N. S. Jan 21 '12 at 22:31
A simple example of a non-null set that does not contain any intervals would also be the set of all irrational numbers. – Dejan Govc Jan 21 '12 at 22:33
up vote 5 down vote accepted

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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Thank you! both you and Arturo have given me perfect answers, but i'll give you the accept as you have less rep :) – Freeman Jan 21 '12 at 22:40
@LHS That's very kind of you, thank you. Glad I could help. : ) – Rudy the Reindeer Jan 21 '12 at 22:41

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.. – Freeman Jan 21 '12 at 22:39

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