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Can anyone provide a sequence of singular (w.r.t. Lebesgue measure) measures $\in\mathcal{M}([0,1])=C[0,1]^*$ converging $weakly^*$ to an absolutely continuous (w.r.t. Lebesgue measure) measure?

J.R.
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checkmath
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    Two small requests: 1. If you re-post a question, either here or on MO please say that you asked the same (or a very closely related one) elsewhere. 2. Please don't introduce new tags, stick to those that are available. – t.b. Mar 10 '12 at 08:52

2 Answers2

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$$\mu_n = \sum_{i=1}^n \frac{1}{n} \delta_{\frac{i}{n}}$$

Nate Eldredge
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If $(x_n)$ is a sequence of points in $[0,1]$, then $x_n$ is equidistributed with respect to the absolutely continuous measure $f(x)dx$ (where $dx$ denotes Lebesgue measure) precisely if the sequence of singular measures $\dfrac{1}{n} \sum_n \delta(x-x_n)$ on $[0,1]$ converges in the weak-$*$ topology to the measure $f(x)dx$.

For example, if $(x_n)$ is the sequence $(\alpha n \bmod 1)$ for an irrational number $\alpha$ then $(x_n)$ is equidistributed with respect to Lebesgue measure $dx$.

KCd
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Matt E
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