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Hi,

I have a question about the circular law. Consider $A_n=[x_{ij}]$ a sequence of random matrices where $x_{ij}$ are iid with mean $0$ and variance $1$. Consider $\lambda_{n,1},\dots,\lambda_{n,n}$ the eigenvalues of $\frac{1}{\sqrt{n}} A_n$ and $\mu_n$ the random measure $\frac{1}{n}\sum_{k=1}^n \delta_{\lambda_{n,k}}$ on $\mathbb{C}$. The circular law (see the paper of Tao-Vu-Krishnapur, 2010, annals of proba) states that for each test function $f:\mathbb{C}\rightarrow \\mathbb{C}$ one has almost surely $$ \lim\limits_{n\rightarrow +\infty} \int_{\mathbb{C}} f(z) d\mu_n = \frac{1}{\pi} \int_{p^2+q^2\leq 1} f(p+iq) dpdq $$

By test function, we mean continuous and compactly supported. Is the same conclusion also true if $f$ is the caracteristic function of a closed subset of $\mathbb{C}$ ? Here is a simple example, consider a closed subset $F\subset \mathbb{C}$ with zero Lebesgue measure, is it true that $$\lim\limits_{n\rightarrow +\infty} \mathbb{P}( \forall k \in [1,n], \lambda_{n,k}\in F ) =0 $$

Thanks,

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    $\begingroup$ By Urysohn's lemma, you can approximate $1_F$ from above in $L^1$ by a sequence of continuous compactly supported functions, $(f_n)$, and below by 0. Applying the result of the TVK paper to the $f_n$'s, you get a slightly weaker conclusion than you're asking for. $\endgroup$
    – Anthony Quas
    Commented Mar 11, 2013 at 6:32
  • $\begingroup$ Which $L^1$ space are you thinking about ? with the Lebesgue measure or $\mu_n$ measures ? So you say that the answer of my question would be affirmative ? $\endgroup$
    – jeanB
    Commented Mar 11, 2013 at 12:16
  • $\begingroup$ The (non-random) $L^1$ space with Lebesgue measure. This will mean the right side of the TVK equality (which is an upper bound) converges to 0. But this only gives that $\mathbb E\mu_{n}(F)\to 0$, which is a weaker conclusion than you asked for. $\endgroup$
    – Anthony Quas
    Commented Mar 11, 2013 at 16:24

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