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Let $X$ be a set, $\mathcal F$ a $\sigma$-field of subsets of $X$, and $\mu$ a probability measure on $X$.

Given random variables $f,g\colon X\rightarrow\mathbb{R}$, suppose I want to show that $\mu_f=\mu_g$. Does it suffice to show that $\mu_f(A)=\mu_g(A)$ for all sets $A$ that is a union of disjoint intervals?

Kunal
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    Yes. Hint: consider the class of Borel subsets $A$ of $\mathbb R$ such that $\mu_f(A)=\mu_g(A)$. – Did Nov 19 '13 at 10:52

1 Answers1

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The class of finite disjoint union of intervals (open/closed/other type) forms an algebra: it contains the empty set, it is stable under complementation and under finite intersections.

Then approximate.

Davide Giraudo
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