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Can anyone say how to relate Stone-Weierstrass approximation theorem in this problem? Given that $\mu$ and $\nu$ are two probability measure on the real line, if for all bounded continuous functions $f$, $\int f d \mu = \int f d \nu$, then $\mu=\nu$?

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    elinor, I _strongly encourage you_ to start accepting answers to your questions. This is a basic courtesy to show that you appreciate the work they've done, for free, to assist you, a total stranger.2011-03-07

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I believe the intended solution is to approximate the indicator function of a measurable set (which is not continuous!) by bounded continuous functions. A measure is determined by the measure it assigns to measurable sets, so if you can establish $\mu(A) = \nu(A)$ for every interval $A \subseteq \mathbb{R}$ (or indeed, for any other $\pi$-system generating the $\sigma$-algebra in question), that proves that $\mu = \nu$.