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Just read the last two lines of pg2 of the notes, why the last integral is supported on $\{|f(x)| > \alpha\}$ but not $\{|f(x)|\geq\alpha\}$? Since we can define $f(x,\alpha)=1.\alpha^{p-1}$ which is supported on the set $\{(x,\alpha)\in (R^n,R)|0\leq\alpha\leq f(x)\}$ , fubini theorem should come in here .

Can anyone give me a more detailed explanation of the proof by using fubini ? Thanks in advance .

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Since the definition of $\lambda_f$ contains $>$, it is correct to integrate on $\{(x,\alpha)\mid 0 < \alpha < f(x)\}$. – Siminore Sep 28 '12 at 13:30
@Siminore but why not on the closed one ? Since in the previous integral it is clear that $\alpha^{p-1}$ is integrated on the closed interval [0,f(x)] – teshile Sep 28 '12 at 13:55
In Lebesgue's theory, $\int_{[a,b]} = \int_{(a,b)}$. – Siminore Sep 28 '12 at 14:21

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