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Let $(X,M,\mu)$ is measure space

$f$ is non negative measurable function .

E is measurable set in M .

then $\int_E f=\int \chi_E f$

My Attempt: $\int f=\sup_{0\leq \phi \leq f }{\int \phi }$ where $\phi$ is simple function

$\int_E f=\sup_{0\leq \phi \leq f }{\int \phi }=\sup_{0\leq \chi_E\phi \leq \chi_E f }{\int \phi }$ =$\int \chi_E f$

I have doubt that is above correct? Actually I had proved charactersitics function of type $\chi_E\phi $ but definition required supremum over all possible simple function

Any Help will be appreciated

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  • $\begingroup$ What is $\int_{XE}f$? $\endgroup$ – herb steinberg May 31 '19 at 17:00
  • $\begingroup$ What is your definition of $\int_E f$? The usual definition is $\int_E f=\int\chi_E f$. $\endgroup$ – David C. Ullrich May 31 '19 at 17:16
  • $\begingroup$ Dear Sir , I had mentioned definition in my attempt: $\endgroup$ – MathLover May 31 '19 at 17:17
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    $\begingroup$ $\int_E f=\sup_{0\leq \phi \leq f }{\int_E \phi }$ $\endgroup$ – MathLover May 31 '19 at 17:18
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Note that $\psi$ is a simple function such that $0 \leqslant \psi \leqslant f\chi_E$ on $X$ if and only if $\psi = \phi\chi_E$ where $\phi $ is a simple function such that $0 \leqslant \phi \leqslant f$ on $E$.

Thus,

$$\int_E f = \sup_{0 \leqslant \phi \leqslant f}\int_E\phi = \sup_{0 \leqslant \phi \leqslant f}\int_X\phi\chi_E = \sup_{0 \leqslant \phi\chi_E \leqslant f\chi_E}\int_X\phi\chi_E = \sup_{0 \leqslant \psi \leqslant f\chi_E}\int_X\psi = \int_X f\chi_E$$

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