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Suppose $\mu_n$ is a sequence of probability measures on some compact space with the Borel sigma-algebra. Define a new function $\mu$ by the equation $\mu(A)=\sum_{n=1}^\infty 2^{-n}\mu_n(A)$ for every Borel set A. Is $\mu$ necessarily a measure?

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You may want to see the general context this can be showed in, e.g. at ProofWiki. Since each $\mu_i$ is a probability measure, so is $\mu$ (since $\sum_n 2^{-n} = 1$) – Lord_Farin Oct 20 '12 at 12:30
up vote 3 down vote accepted

$\mu$ is a set function, and $\mu(\emptyset)=\sum_{n\geq 1}2^{—n}\mu_n(\emptyset)=0$. Let $\{A_k\}_{k\geq 1}$ a sequence of pairwise disjoint measurable sets. As $\mu_n(A_k)\geq 0$ for each $n$ and $k$, we can switch the two sums: $$\small\mu\left(\bigcup_{k\geq 1}A_k\right)=\sum_{n\geq 1}2^{—n}\mu_n\left(\bigcup_{k\geq 1}A_k\right)=\sum_{n\geq 1}2^{-n}\sum_{k\geq 1}\mu_n(A_k)= \sum_{k\geq 1}\sum_{n\geq 1}2^{—n}\mu_n(A_k)=\sum_{k\geq 1}\mu(A_k).$$ As $\mu_n(\Omega)=1$ for each $n$ and $\sum_{n\geq 1}2^{—n}=1$, $\mu$ is a probability measure.

Compactness and the fact that we worked with a Borel $\sigma$-algebra wasn't used.

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