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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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What is $\mu\{x \in \mathbb{R}\,:\,f(x) \neq g(x)\}$ and what must be true if it is zero? –  t.b. Oct 25 '11 at 15:43
    
That measure would be zero, as it is where the "a.e." part doesn't cover. What must be true is that there is a subset of the sequence $\{x_n\}$ where $f\neq g$ and this set is measure zero - so a countable set for instance. This would also then be Borel, which is the algebra, which is why I thought it might be a complete measure. –  nate Oct 25 '11 at 15:47
    
The set $B = \{x_n\,:\,n \in \mathbb{N}\}$ and all its subsets are certainly Borel sets (that is because they are countable). So $\{x \in \mathbb{R}\,:\,f(x) \neq g(x)\} \subset \mathbb{R} \smallsetminus B$. But, as you said, $\mu(X \smallsetminus B) = 0$, so $\{x \in \mathbb{R}\,:\,f(x) \neq g(x)\}$ is contained in a null set, and this is precisely the definition of $f = g$ $\mu$-a.e. However, the measure $\mu$ is not complete, as there are non-Borel subsets of the null-set $\mathbb{R} \smallsetminus B$; your notation indicates that you consider $\mu$ only on the Borel sets of $\mathbb R$ –  t.b. Oct 25 '11 at 15:55
    
So for me to be clear... the points at which $f(x) \neq g(x)$ ($X\backslash B$) do not exist in the sequence, $B$. So then this would clearly imply $f(x_n) = g(x_n)$ for each $n$. But going the other way, does the fact that there are non-Borel subsets of $\mathbb{R}\backslash B$, along with $f(x_n) = g(x_n)$, imply $f=g$ at $(\mathcal{B}(\mathbb{R}),\mu)$-a.e.? –  nate Oct 25 '11 at 16:09
1  
The definition of $f = g$ $(\mathcal{B}(\mathbb{R}),\mu)$-a.e. is: there exists a $\mu$-null set $N \in \mathcal{B}(\mathbb{R})$ such that $\{x \in X\,:\,f(x) \neq g(x)\} \subset N$. It is not required that this set $\{x \in X\,:\,f(x) \neq g(x)\}$ be measurable itself. Note also that you don't have any measurability condition on $f$ and $g$. –  t.b. Oct 25 '11 at 16:13

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