Is the countable union of measure-zero sets zero?

Let $E=\cup_{i=1}^\infty A_i$ where the measure of $A_i$ is zero. How can I conclude that $E$ has measure zero?

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If your measure $\mu$ is sub sigma additive you can surely say that
\[ \mu(E)\leq \sum_{i=1}^\infty \mu (A_i)=0\]
If it’s not the statement doesn’t need to be true.

Hint: Assuming the measure is Lebesgue measure, for each $A_i$, you can find an open set $O_i$ such that $A_i\subseteq O_i$ and the measure of $O_i$ is less than $\varepsilon/2^i$. If not, as long as the measure is subadditive, you can follow Dominic Michaelis’ approach.

It follows by sigma-aditivity.
$$m\bigg(\bigcup_{i\in\mathbb {N}}A_{i}\bigg)\leq\sum_{i\in\mathbb{N}}m(A_{i})=0$$