Let $(X,\mathcal A, \mu)$ be a measure space where $X$ is topological space and $\mu$ is a Borel measure on $\mathcal A.$ Let $f : X \longrightarrow \mathbb C$ be a continuous function which is zero $\mu$ a.e. Does it necessarily imply that $f = 0$ identically on $X\ $?
If $X$ is a topological group and $\mu$ is a Haar measure on the Borel-$\sigma$-algebra of subsets of $X$ then I know that it is true. What about any arbitrary topological space $X\ $? Any help in this regard would be much appreciated.
Thanks!
Not always! Take $X = \{a, b, c\}$ with the discrete topology, and let $\mu$ be the measure $\mu(\{a\}) = 1$ and $\mu(\{b, c\}) = 0$ (this uniquely determines a measure on the power set of $X$). Then $f(a) = 0, f(b) = f(c) = 17$ is a counterexample to the statement. In general, measures can be very wild.