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

  • $\begingroup$ 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
  • 2
    $\begingroup$ $\int_E f=\sup_{0\leq \phi \leq f }{\int_E \phi }$ $\endgroup$ – MathLover May 31 '19 at 17:18

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 $E$.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.