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I want to ask if it is true in general topological space that the countable union of sets of measure $0$ has $0$ measure?

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It is true in general measure spaces as a consequence of countable additivity. –  Michael Greinecker Apr 15 '12 at 21:43
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On the other hand, a countable union of meager sets is meager, in any topological space (meager means it is a countable union of nowhere dense subsets). Meager sets are the usual notion of "negligible" in the context of topological spaces, while "contained in a set of measure zero" is the usual notion of "negligible" in measure spaces. –  Arturo Magidin Apr 15 '12 at 23:19
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A general topological space doesn't have a notion of "measure 0".

The statement you ask about is true in a general measure space. By definition, measures are required to be countably additive. This means that $$ m\left(\bigcup_{i=1}^\infty S_i\right) \;=\; \sum_{i=1}^\infty m(S_i) $$ for any countable disjoint collection of measurable sets $\{S_i\}_{i=1}^\infty$. As a consequence, if $\{S_i\}_{i=1}^\infty$ is any countable collection of measurable sets (not necessarily disjoint), then $$ m\left(\bigcup_{i=1}^\infty S_i\right) \;\leq\; \sum_{i=1}^\infty m(S_i) $$ In particular, if each $S_i$ has measure $0$, then the union must also have measure $0$.

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"Measure" doesn't make sense in a general topological space. In a general measure space, a countable union of measure-zero sets always has measure zero, by the countable subadditivity of measures.

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