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Is the following proof valid?

Let $(X, \mathcal{B})$ be a measurable space and suppose $\lambda_n$ is a sequence of finite measures s.t. $\lambda = \sum_{n=1}^\infty \lambda_n$ is finite. I want to show that if $\mu$ is a measure s.t. $\lambda_n\,\bot\,\,\mu$ for all $n$, then $\lambda\,\bot\,\, \mu$ as well.

Note that I'm using Royden's notion of a mutual singular measure: that is, $\lambda \,\bot\,\, \mu$ iff $\exists A, B \in \mathcal{B}$ s.t. $A \cup B = X$, $A \cap B = \emptyset$ and $\mu(A) = \lambda(B) = 0$.

  1. Consider that for each $n \in \mathbb{N}$, $\lambda_n\,\bot\,\,\mu$ implies that $\exists A_n, B_n \in \mathcal{B}$ s.t. $\mu(A_n) = \lambda_n (B) = 0$ with $A_n \cup B_n = X$ and $A_n \cap B_n = \emptyset$.

  2. Let $A = \bigcup_{n=1}^\infty A_n$ and $B = A^c$ so that $X = A \cup B$ and $A \cap B = \emptyset$.

  3. I claim that $\mu(A) = 0 = \lambda(B)$.

  4. First note that if we arbitrarily fix $k \in \mathbb{N}$, then $\lambda_k (B) = 0$. This is because $A \supseteq A_k$ implies that $A^c = B \subseteq B_k = A_k^c$ so that by the monotonocity of measures we have $\lambda_k (B) \le \lambda_k (B_n) = 0 \implies \lambda_k (B) = 0$.

  5. Since $\forall k \in \mathbb{N}$, $\lambda_k (B) = 0$, it now follows that $\lambda(B) = \sum_{k=1}^\infty \lambda_k (B) = 0$ as well.

  6. Finally, if we assumed for sake of contradiction that $\mu(A) > 0$, then we would have $\mu(\bigcup_{n=1}^\infty A_n) > 0$. Since $\mu$ is countably sub-additive relative to any collection of measurable subsets, we have $0 < \mu(\bigcup_{n=1}^\infty A_n) \le \sum_{n=1}^\infty \mu(A_n) = \sum_{n=1}^\infty 0 = 0$ which is clearly absurd.

  7. Then with $\mu(A) = 0 = \lambda(B)$ with $A \cup B = X$ and $A \cap B = \emptyset$, it follows that $\mu\,\, \bot \, \lambda$ as desired.

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Your solution looks good to me. However, you don't need to prove (6) with contradiction. You can appeal directly to sub-additivity to get $\mu(A) = \mu(\bigcup_1^\infty A_n) \leq \sum_1^\infty \mu(A_n) = 0$. –  Ler Dec 18 '12 at 20:22

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