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I am stuck on this problem:

Let $\{x_n\}$ be a sequence of real numbers, and define $\mu := \sum_n \delta_{x_n}$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Show that functions $f,g: \mathbb{R} \to \mathbb{R}$ agree $\mu$-a.e. iff $f(x_n) = g(x_n)$ holds for each $n$.

So far I have:

$\Rightarrow$ $f=g$ at $(\mathcal{B}(\mathbb{R}),\mu)$-a.e. implies $f(x_n) = g(x_n)$ for each $n$

It is plausible to let $f(x_n) = g(x_n)$ at all $x_n$ in the sequence and let a countable number of additional points in the sequence exist s.t. $f(x_{n+1}) \neq g(x_{n+1}), f(x_{n+2}) \neq g(x_{n+2}),\ldots$ (Is this a complete measure as every countable set of real numbers is a Borel set of measure zero? So the set of zero measure is included in the algebra?) For some countable number of points in the sequence, the relation fails.

$\Leftarrow$ $f(x_n) = g(x_n)$ holds for each $n$ implies $f=g$ at $(\mathcal{B}(\mathbb{R}),\mu)$-a.e.

?

Any help would be appreciated.

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    Okay, I think I get it pretty clearly now, especially clearing up the complete part - and that f,g have no measurability conditions. Thanks!2011-10-25

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