I want to ask if it is true in general topological space that the countable union of sets of measure $0$ has $0$ measure?
Countable union of measure $0$ sets has measure $0$
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2On 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. – 2012-04-15
2 Answers
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 $0$.
"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.