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Proof of [tex]\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A[/tex]

  1. Jun 1, 2012 #1
    [tex]\text{ Let }f:\mathbb{R}\to [0,\infty] \text{ be a measurable function and }A\subset \mathbb{R}[/tex]Then, show that
    \begin{equation}
    \int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A \tag{1}
    \end{equation}
    [tex]\text{ where ${1}_A$ is the characteristic function of $A$ defined as }[/tex]
    \begin{equation}
    {1}_A(x)=\begin{cases}1 & \text{if $x\in A$,}
    \\
    0 &\text{if $x\notin A$.}
    \end{cases} \tag{2}
    \end{equation}
    [tex]\text{ and $\int\limits_{A}f$ is the Lebesgue integral of $f$ on $A$ defined as:}[/tex]
    \begin{equation}
    \int\limits_{A}f=\sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\} \tag{3}
    \end{equation}

    I can easily prove this property for simple functions so take this for granted:

    \begin{equation}
    \int\limits_{A}s=\int\limits_{\mathbb{R}}s{1}_A \tag{4}
    \end{equation}

    [tex]\text{where $s:\mathbb{R}\to [0,\infty]$ is a simple function. Thus to prove (1) we need to show that:}[/tex]
    \begin{gather}
    \sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\}=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f{1}_A\text{ and }s\text{ is simple}\right\}\notag\\
    \sup\left\{\int\limits_{\mathbb{R}}s{1}_A:0\le s\le f\text{ and }s\text{ is simple}\right\}=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f{1}_A\text{ and }s\text{ is simple}\right\} \tag{5}
    \end{gather}

    My question is how do we prove (5)?
     
  2. jcsd
  3. Jun 1, 2012 #2
  4. Jun 2, 2012 #3
    The statement can be easily shown via the MCT.
     
  5. Jun 4, 2012 #4
    Specifically, I was thinking something like this:

    the statement is obvious for simple functions. Let f be some nonnegative measurable function on [itex]\mathbb{R}[/itex]. Construct a sequence of simple functions [itex]\{s_n\}_{n=1}^{\infty} [/itex] so that [itex] s_n(x) \uparrow f(x) [/itex] as [itex] n \to \infty [/itex]. Then we also have that [itex](s_n 1_A)(x) \uparrow (f 1_A)(x) [/itex].

    By the monotone convergence theorem, we have that [itex] \int_A s_n \to \int_A f [/itex] as [itex] n \to \infty [/itex] and that [itex] \int_{\mathbb{R}} s_n 1_A \to \int_{\mathbb{R}} f 1_A [/itex] as [itex] n \to \infty [/itex].

    But we know that for each n, [itex] \int_{\mathbb{R}} s_n 1_A = \int_A s_n [/itex]. So we have that [itex] \int_A s_n \to \int_{\mathbb{R}} f 1_A [/itex]. But limits of real sequences are unique, so it follows that [itex] \int_A f = \int_{\mathbb{R}} f 1_A [/itex].
     
  6. Jun 5, 2012 #5
    Let L(f) = ∫Af-∫ℝf1A. We note L(s) = 0 for all step maps. Since L is continuous in the L1 norm and step maps are dense in L1 we have that L(f) = 0 for all f.
     
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