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Given: $P=\sum_{n=1}^{\infty}2^{-n}P_n$

I am trying to show that it is a probability measure, and further, that:

$\int_{\Omega}XdP=\sum_{n=1}^{\infty}2^{-n}\int_{\Omega}XdP_n$ for any non-negative or bounded random variable $X$

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1 Answer 1

up vote 3 down vote accepted

I'm assuming all the $P_n$'s are measures on the same sigma algebra. To show that $P$ is a measure, you need to show additivity: let $V_k$ be a collection of disjoint sets and let $V=\bigcup V_k$. Then, $$P(V)=\sum_n 2^{-n}P_n(V)=\sum_n\sum_k2^{-n}P_n(V_k)=\sum_k\sum_n2^{-n}P_n(V_k)=\sum_k P(V_k)$$ where the exchange of the sums is due to the monotone convergence theorem applied to sums.

The integral equality is valid for simple functions by what was just shown. You can find a sequence of simple functions that approximate X, i.e. $X_n\to X$ pointwise and monotonically on the positive and negative domains of $X$. Then, $$\int_\Omega XdP=\lim_k\int_\Omega X_k dP=\lim_k\sum_n 2^{-n}\int_\Omega X_kdP_n=\sum_n 2^{-n}\lim_k\int_\Omega X_kdP_n=\\ =\sum_n 2^{-n}\int_\Omega XdP_n$$ where the exchange of sum and integral is due monotone convergence when X is non-negative, and dominated convergence when X is bounded.

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