# Proving if $A$ is a null set and $B$ is a countable set then $A+B$ is null

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

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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
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## 2 Answers

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. : ) – Matt N. Jan 21 '12 at 22:41

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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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