4
$\begingroup$

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.

  • 0
    Though notations are different, and is also for simple functions, similar proof is sketched here. [MathCS](http://www.mathcs.org/analysis/reals/integ/answers/simpint5.html)2012-12-31

2 Answers 2

1

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 $f$."

  • 0
    Thanks. I will try this one, too !2013-01-01
  • 0
    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 :)2013-01-01
1

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)

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.

  • 0
    I am still quite unsure what to do :D2012-12-31
  • 0
    Is it OK now ? I included the definition of integration that I use2012-12-31
  • 0
    Thanks for that. Assume I can prove that, is your definition of $\int_X f$ equal to the definition of Tao which is $$ \int_X f := \sup \left \{ \int_X s \mid 0 \leq s \leq f \text{ and $s$ simple} \right \}$$2012-12-31
  • 0
    I believe they are equivalent2012-12-31
  • 0
    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 $$\int_X f := \sup \left \{ \int_X s \mid 0 \leq s \leq f \text{ and $s$ simple} \right \} = \inf \left \{ \int_X s \mid 0 \leq f \leq s \text{ and $s$ simple} \right \}$$.2012-12-31
  • 0
    I have done it now as follows : First I prove that, if $f(x) \leq g(x)$ a.e. then $\int_\Omega f \leq \int_\Omega g$. The original claim follows then easily because $f \leq g$ a.e. and $g \leq f$ a.e. and so the integrals do match.2012-12-31
  • 0
    This is true. But you also have to show that $f\leq g$ a.e implies that $\int_{X} f\ d\mu\leq\int_{X} g\ d\mu$2012-12-31
  • 0
    I have proven this. But I am unsure about a little thing. Assume $s' = s \chi_{\Omega \backslash E}$ where $m(E) = 0$ and $s$ is simple. Is it then true, that $$ \int_\Omega s' = \int_\Omega s \chi_{E^c} = \int_{E^c} s = \int_\Omega s - \int_E s$$ ?2012-12-31
  • 0
    And hopeful $\int_E s = 0$ ?2012-12-31
  • 0
    $$\int_{\Omega}s'=\int_{\Omega}s -\int_{\Omega\cap E}s$$2012-12-31
  • 0
    What you did on your last comment uses what you aren't allowed to use. What you have is actually $$\int_{E^c}s=\int_\Omega s(1-\chi_{E})=\int_\Omega s - \int_{E}s$$. Thus to do the step you have done, you are using what you said we can't assume.2012-12-31
  • 0
    You are referencing to the last post of "Amr" ?2013-01-01
  • 0
    Ah yes, sorry about that.2013-01-01