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Let $(\mathbb{R}, \mathcal{B}, \mathbb{P})$ be a probability space. I want to show that $\mathbb{P}$ can be written as $\mathbb{P} = \mu + \nu$, where $\mu$ is a continuous measure (no atoms) and $\nu$ an atomic measure ($\nu = \sum_i \epsilon_i \delta_{x_i}$).

I think first one has to show, that $\mathbb{P}$ has only countable many atoms $x_i$. Then, that $\mathbb{P} - \nu = \mu$ is a measure.

The second part should not be to hard. But how to do the first? If there are uncountable many $x_i \in \mathbb{R}$ with $\mathbb{P}(x_i) >0$, I guess $\mathbb{P}(\mathbb{R})=1$ is not possible anymore, but how to show that?

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up vote 2 down vote accepted

Suppose that there are uncountably many such $x_i$, indexed by $i\in I$. Then $$ 1=\mathsf P(\mathbb R)\geq \sum\limits_{J}\mathsf P(x_j) $$ where $J\subseteq I$ is any finite set. You can show that if $\mathsf P(x_i)>0$ for all $i\in I$ which is uncountable, then for any $E\geq 0$ there is $J(E)$ such that $\sum\limits_{J(E)}\mathsf P(x_j)\geq E$. Then take $E = 2$ and you're done.

Let us show that if $p_i >0$ for uncountably many $i\in I$ then there exists a finite $J\subset I$ such that $$ \sum_J p_i>2. $$ Let us put $I_n = \{i\in I:p_i\geq 1/n\}$. Suppose that each $I_n$ is finite, then $I = \bigcup\limits_{n=1}^\infty I_n$ is countable as a countable union of finite set. But in our case $I$ is uncountable, hence as a contradiction, there is $N$ such that $I_N$ is infinite. Take any finite $J\subseteq I$ of cardinality $\# J\geq 2N$, then $$ \sum\limits_J p_i\geq \sum\limits_J \frac1N\geq 2. $$

Above, $2$ can be replaced with an arbitrary positive integer number.

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@Haatschii Please tell me if you need some hints to show that $$ \sum\limits_J \mathsf P(x_j) $$ can be made as large as we want. –  Ilya Jul 16 '12 at 15:52
    
yes, this is in fact the point where I get stuck. A hint would be welcome. –  Haatschii Jul 16 '12 at 16:09
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@Haatschii: updated –  Ilya Jul 16 '12 at 16:34
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