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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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closed as off-topic by Bill Johnson, Noah Stein, Pietro Majer, R W, Yemon Choi May 3 at 19:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Bill Johnson, Noah Stein, Pietro Majer, R W, Yemon Choi
If this question can be reworded to fit the rules in the help center, please edit the question.

    
Disclosure: if we still had the option to "close as no longer relevant", then that is what I would have chosen –  Yemon Choi May 3 at 19:22

3 Answers 3

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).

Ayman

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

This question has been answered but it is part of a larger picture which might be of interest to the OP. If $S$ is a set, then $\Lambda(S)$, the set of formal linear combinations of elements of $S$, is a vector space, the free vector space over $S$. If $S$ has a (topological or analytic) structure, then this can be used to provide the vector space with a suitable locally convex topology. Thus if $S$ is a compact interval, or indeed a compact space $K$, then we give $\Lambda(K)$ the topology defined by all seminorms whose restrictions to $K$ are continuous. This space will not usually be complete so we complete it to get $\tilde \Lambda(K)$. Then

1) $\Lambda(K)$ is dense in the latter;

2) this space has the universal property that each continuous mapping from $K$ into a Banach space has a unique lifting to a continuous linear mapping on $\tilde \Lambda(K)$;

3) the dual is $C(K)$;

4) it can be naturally identified with space of Radon measures on $K$ provided with a topology which is not identical with the weak $\ast$ topology but has the same convergent sequences and the same dual but with the, in our opinion significant, advantage that it is complete.

An important bonus is that this construction has a vast array of interesting variants. Thus one can consider bounded continuous functions on a completely regular space, smooth functions on a manifold (compact or not, with or without boundary), holomorphic functions on complex manifolds, bounded, uniformly continuous functions on uniform spaces, to get spaces of bounded Radon measures, distributions, analytic functionals, uniform measures to name but a few of the interesting possibilities.

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An elementary constructive proof is shown in Example 8.1.6 (i) in Bogachev, V. I., Measure theory Vol. II.

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