# Inner Regularity of Lim-Inf of Measures

Suppose I have a sequence of inner regular measures, is it true that the lim-inf of the sequence an inner regular measure?

I will be more specific for the context of my problem.

Let $\Omega_l$ be such that $\Omega_l \subset \Omega_{l+1}$ and $\cup_l \Omega_l = \Omega$, furthermore $\mu(\Omega_l) , \mu(\Omega) < \infty$, $\mu$ denotes the Lebesgue measure. Let $u_{\varepsilon} \geq 0$ be a sequence of bounded integrable functions, then does $$\lim_l \liminf_{\varepsilon} \int_{\Omega_l} u_{\varepsilon}(x)dx = \liminf_{\varepsilon} \int_{\Omega} u_{\varepsilon}(x)dx ?$$

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Borel measure on $\mathbb R^d$. –  dcs24 Nov 27 '12 at 17:19
I have a sequence of measures $\mu_n$ for $n \in \mathbb N$. I take the lim inf pointwise, that is for each Borel set $U$ $\liminf_{n \rightarrow \infty} \mu_n(U)$. –  dcs24 Nov 28 '12 at 12:57
Updated the question to be more specific. –  dcs24 Nov 28 '12 at 13:54
You may also assume that the lim inf is finite. –  dcs24 Nov 28 '12 at 13:56
No problem, now I will think on it. –  Davide Giraudo Nov 28 '12 at 17:12

Assume that $\mu(\Omega_l)>\mu(\Omega_{l-1})$, and let $u_n:=\frac{\chi_{\Omega_n\setminus\Omega_{n-1}}}{\mu(\Omega_n)-\mu(\Omega_{n-1})}$. It's a bounded non-negative function, and $\int_{\Omega}u_nd\mu=1$ for all $n$, so the RHS is $1$. For a fixed $l$, $$\liminf_{n\to +\infty}\int_{\Omega_l}u_nd\mu=0.$$