Measurable Spaces – Problem (49/365)

Let \mu be the Lebesque-Stieltjes measure generated by a continuous function. Show that if the set A is at most countable, then \mu(A) = 0. A Lebesque-Stieltjes measure is a countably additive measure and is given by a generalized distribution function G(x) such that \mu([a,b)) = G(b) - G(a) that is continuous on the right. So,

\displaystyle  A = \cup_{i=1}^\infty \{ a_i \} \\ \mu(A) = \sum_{i=1}^\infty \mu(\{ a_i \}) \text{ by countable additivity}\\ \mu(\{ a_i \}) = \lim_{n \rightarrow \infty} \mu [a_i, a_i + \frac{1}{n}) \\ = \lim_{n \rightarrow \infty} G(a_i + \frac{1}{n}) - G(a_i) \\ = 0 \\ \text{Thus } \mu(A) = 0

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