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Let $f,g : \Omega \subseteq \mathbb R^n \rightarrow [0,+\infty]$ be measurable functions with $f(x) = g(x)$ a.e. . Then I have to show that $\int_\Omega f = \int_\Omega g$.

I may not assume that $\int_\Omega (f+g) = \int_\Omega f + \int_\Omega g$.

This task is from Tao Proposition 19.2.6.

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  • $\begingroup$ Though notations are different, and is also for simple functions, similar proof is sketched here. MathCS $\endgroup$ – 007resu Dec 31 '12 at 13:49
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You can show that the set $$ \{ \int_\Omega s : 0\leq s\leq f\}=\{\int_\Omega s : 0\leq s\leq g\}$$ by using a double inclusion argument and the fact stated in the comments of the original question that simple functions which equal almost everywhere have the same integral. Therefore their supremums are equal which respectively equal the integrals you want.

The sets I mentioned above are equivalent to what Tao calls "simple functions that minorize $f$."

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  • $\begingroup$ Thanks. I will try this one, too ! $\endgroup$ – user42761 Jan 1 '13 at 13:50
  • $\begingroup$ I choose this as answer because it is easier. And in fact I only have to show that if $s = s \chi_{E}$ such that $m(E^c) = 0$ then $\int_\Omega s = \int_\Omega s \chi_E$. This is easy to prove. Thanks :) $\endgroup$ – user42761 Jan 1 '13 at 14:07
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The definition of integration that I am using: The predicate $CSM(p)$ is a shorthand for $p$ having a countable range and $p$ is a measurable function. Let $p:X\rightarrow [0,\infty]$ be a function such that $CSM(p)$. We will define the integral of $p$ to be: $$\int_X p\ d\mu=\sum_{a\in p(X)}a\mu(p^{-1}(\{a\}))$$ Now let $f:X\rightarrow [0,\infty]$ be any measurable function, we define the upper and lower integrals of $f$ as:

$$\int_{X}^{*}f\ d\mu=\inf\{\int_{X} p \ d\mu|p:X\rightarrow[0,\infty],CSM(p),\mu(\{x\in X|f(x)>p(x)\}=0\}$$ $$\int_{*X}^{}f\ d\mu=\sup\{\int_{X} q \ d\mu|q:X\rightarrow[0,\infty],CSM(q),\mu(\{x\in X|f(x)<q(x)\}=0\}$$

Finally, we say that the integral of $f$ exists iff $\int_{X}^{*} f\ d\mu=\int_{*X}^{}f\ d\mu$ and we denote it $\int_X f\ d\mu$.

To prove your theorem Verify that if $f=g$ $\mu.a.e$ then:

1) For all measurable functions $p$ , $f\leq p$ a.e. iff $g\leq p$ a.e.

2) For all measurable functions $q$ , $f\geq q$ a.e. iff $g\geq q$ a.e.

Finally deduce that the upper and lower integrals of $f,g$ are equal.

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  • $\begingroup$ I am still quite unsure what to do :D $\endgroup$ – user42761 Dec 31 '12 at 13:58
  • $\begingroup$ Is it OK now ? I included the definition of integration that I use $\endgroup$ – Amr Dec 31 '12 at 14:10
  • $\begingroup$ Thanks for that. Assume I can prove that, is your definition of $\int_X f$ equal to the definition of Tao which is $$ \int_X f := \sup \left \{ \int_X s \mid 0 \leq s \leq f \text{ and $s$ simple} \right \}$$ $\endgroup$ – user42761 Dec 31 '12 at 14:16
  • $\begingroup$ I believe they are equivalent $\endgroup$ – Amr Dec 31 '12 at 14:18
  • $\begingroup$ I guess, too because simple funtions are a special case of your functions which you use in the definition of upper and lower intergral. And Tao says too, that $$\int_X f := \sup \left \{ \int_X s \mid 0 \leq s \leq f \text{ and $s$ simple} \right \} = \inf \left \{ \int_X s \mid 0 \leq f \leq s \text{ and $s$ simple} \right \}$$. $\endgroup$ – user42761 Dec 31 '12 at 14:20

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