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I am studying the proof of a theorem and in a part of the proof I have the following situation:

Let $u : \Omega \rightarrow R$ a nonnegative measurable function, with $\Omega$ open and bounded. Consider $c>0$ a constant. Write $v_{n} = log( u + 1/n )$. Supose that $log (u + 1/n) \in H^{1,p}_{loc} (\Omega)$.

I need to show that $lim_{n \rightarrow \infty} \int_{B} e^{-c_1 v_n} = \int_{B} u^{-c_1}$, where B is a open ball with $2B \subset \Omega$.

I think the dominated convergence theorem can help.. Someone can give a hint to prove what i said?

Thankyou!

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  • $\begingroup$ in the set $\{ u < 1\}$ , we have $e^{-c_1 log (u + 1/n)} \leq e^{c_1 log(1/u)}$ . I dont know if this help... I said this because may can help separate the integral where u > 1 and in u< 1 $\endgroup$ – math student May 12 '13 at 2:38
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I don't think we need anything about Sobolev spaces here. The function $u^{-c_1}$ takes values in $[0,+\infty]$, so $\int_B u^{-c_1}$ is defined as a number in $[0,+\infty]$. The functions $f_n:=\exp(-c_1v_n)$ are nonnegative and form an increasing sequence: $f_{n+1}\ge f_n$. Also, $f_n\to u^{-c_1}$ pointwise. By the Lebesgue monotone convergence theorem we have $\lim \int_B f_n=\int \lim f_n$.

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