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Let $I$ be a compact interval and $\mathcal{M}(I)$ the space of (signed) Borel measures. We equip it with the weak topology, i.e. a sequence $\mu_n$ converges to zero if and only if $$ \left|\int_I f(x) \mathrm{d}\mu_n(x)\right| \longrightarrow 0$$ for all $f \in C(I)$.

Now the question is the following: Let $V \subset \mathcal{M}(I)$ be the vectorspace of all finite linear combinations of Dirac measures supported at different points in $I$. Is $V$ dense in $\mathcal{M}(I)$?

For example if $I = [0,1]$, the sequence $$ \mu_n = \frac{1}{N}\sum_{j=1}^N \delta_{j/N},$$ $\delta_{j/N}$ being the Dirac measure supported at $j/N$, weak*-converges to the Lebesgue measure as $\mu_n$ is just the approximation by Riemann sums. Hence one can easily get all measures that are absolutely continuous w.r.t. the Lebesgue measure.

However, there are more measures (singular measures) that are neither point measures nor Lebesgue measures and I don't have an idea how to reach those.

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up vote 6 down vote accepted

Equipped with the mentionned weak($-\star$) topology, am I wrong or the set of continuous linear forms on $\mathcal{M}(I)$ is precisely given by $C(I)$ ?

Then by the classical use of Hahn-Banach theorem, your vectorspace $V$ if dense if and only if the only continuous linear form of $\mathcal{M}(I)-w\star$ vanishing on $V$ is $0$.

But since $V$ contains all Dirac masses, it seems clear that a the linear form associated with a continuous function $f\in C(I)$ vanishes on $V$ if and only if $f=0$ !

So I would say that $V$ is indeed dense (but only for the weak-$\star$ topology of course).


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What an easy argument! Thanks! –  Kofi May 22 '13 at 10:41
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