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Math Help - measure theory question

  1. #1
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    measure theory question

    So I have a measurable, non-negative function f. I am to show that the integral of f with respect to x is zero if and only if the set of xsthat make f(x) > 0 is a set of measure zero. To show the first half of this implication, I am using a contrapositive argument. My classmate suggested that I use simple functions to approximate the integral below, but I'm not sure how to do this. Any thoughts?

    rogerpodger
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  2. #2
    Moo
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    Hello,

    Here is a proof..

    \int f ~ d \mu=0 \quad \Leftrightarrow \quad \mu (\{f > 0\})=0

    Let (\varphi_n) be an increasing positive sequence that converges to f, and where \forall n, ~ \varphi_n is a simple function.
    We can then say that \int_E \varphi_n ~ d \mu=\sum_{\alpha \in \varphi_n(E)} \alpha ~ \mu (\{\varphi_n=\alpha\})

    Note that \int_E \varphi_n ~ d \mu=0 \quad \Leftrightarrow \quad \mu (\{\varphi_n \neq 0\})=0 (this is very simple to prove with the above formula of \int_E \varphi_n ~ d \mu)
    Also note that saying \neq 0 is equivalent to saying >0 since we work on positive functions.

    __________________________________________
    \underline \Rightarrow \quad :

    \int f ~d \mu =0 \Rightarrow \forall n, \int \varphi_n ~d \mu=0 (because 0 \leqslant \varphi_n \leqslant f)

    Therefore, \forall n, ~ \mu(\{\varphi_n > 0\})=0

    But \varphi_n is an increasing sequence. Hence \{\varphi_n > 0 \} is an increasing set.
    \underbrace{\mu(\{\varphi_n > 0 \})}_{=0 ~\forall n} \longrightarrow \mu(\{f > 0\})

    \Longrightarrow \mu(\{f > 0\})=0



    \underline \Leftarrow \quad :

    \mu(\{f > 0\})=0

    Since \{\varphi_n > 0 \} \subseteq \{f > 0\} (because remember that \varphi_n \uparrow f), we have \mu (\{\varphi_n > 0 \}) \leqslant \mu (\{f > 0\})=0 \quad \Rightarrow \quad \mu (\{\varphi_n > 0 \})=0

    Therefore \int \varphi_n ~ d \mu=0 \quad \forall n

    By the monotone convergence theorem, we can conclude :
    \int f ~ d \mu=\lim_n \int \varphi_n ~ d \mu=0


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  3. #3
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    Here's another proof for the more difficult implication (a more direct one, without simple functions):

    Suppose \int f d\mu=0, where is f\geq 0 is measurable.
    Assume by contradiction that \mu(\{x\,|\,f(x)>0\})>0.
    Choose a sequence (\varepsilon_n)_n strictly decreasing to 0. Since \{x\,|\,f(x)>0\}=\bigcup_n \{x\,|\, f(x)>\varepsilon_n\}, where the sets in the union form an increasing sequence, you have: 0<\mu(\{x\,|\,f(x)>0\})=\lim_n \mu(\{x\,|\, f(x)>\varepsilon_n\}). As a consequence, there exists n such that \mu(\{x\,|\, f(x)>\varepsilon_n\})>0, which implies: \int f d\mu\geq \varepsilon_n\cdot \mu(\{x\,|\,f(x)>\varepsilon_n\})>0, in contradiction with the initial assumption.
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