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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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    No problem, now I will think on it.2012-11-28

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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 $l$, $\liminf_{n\to +\infty}\int_{\Omega_l}u_nd\mu=0.$

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    Thank you. Especially for your patience.2012-11-28