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I was reading Brezis enter image description here

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I understand everything except 2 doubt

1) Why u is measurable

2) why $u\chi_n\in L^2(\Sigma)$

I have highlighted text.

Any Help will be appreciated

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Quotient of two measurable functions is still measurable, this follows by the composition of the continuity map $(u,v)\rightarrow u/v$ and $x\rightarrow(f(x),g(x))$ for measurable functions $f,g$.

On the set $\Omega_{n}$, $\theta^{-1}\leq\epsilon_{n}^{-1}$, so $\|u\chi_{\Omega_{n}}\|_{L^{2}}\leq\epsilon_{n}^{-1/2}\|v\|_{L^{2}}$.

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