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Homework Help: Differentiation and Lebesgue integration

  1. Aug 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]g(x) = \int_0^x f(t) dt[/itex], where [itex]f[/itex] is Lebesgue integrable on [itex]\mathbb R[/itex]. Give an [itex]\epsilon - \delta[/itex] proof that [itex]g'(y) = f(y)[/itex] if [itex]y\in (0,\infty)[/itex] is a point of continuity of [itex]f[/itex].


    2. Relevant equations



    3. The attempt at a solution
    I know I need to show that
    [tex]
    f(y) = \lim_{h\to 0} \int_y^{y+h} \frac{1}{h} f(t) dt.
    [/tex]
    My idea was to try to do this in terms of sequences; i.e., to let [itex]\{h_n\}[/itex] be any sequence of real numbers such that [itex]h_n \to 0[/itex], and then to phrase the limit above in terms of a limit as [itex]n\to \infty[/itex]. I had then planned to use something like the dominated convergence theorem. But I don't have any idea how to make use of the hypothesis that [itex]f[/itex] is continuous at [itex]y[/itex], so I'm not sure if this is the right approach.
     
  2. jcsd
  3. Aug 11, 2010 #2

    lanedance

    User Avatar
    Homework Helper

    could you look at it as follows:
    [tex]f(y) = \lim_{h\to 0} \frac{1}{h} \int_y^{y+h} f(t) dt[/tex]

    then use the conitinuity of f to show that as h gets small
    [tex] \int_y^{y+h} f(t) dt \approx f(t).h[/tex]
     
  4. Aug 15, 2010 #3
    Ok. I think I know how to do this. Here is a complete solution: Let [itex]f \in L^1(\mathbb R)[/itex]. We want to show that, given any [itex]\epsilon > 0[/itex], we can find a [itex]\delta > 0[/itex] such that
    [tex]
    \left| \frac{1}{h} \int_y^{y+h} f - f(y) \right| < \epsilon
    [/tex]
    whenever [itex]|h| < \delta[/itex]. Since [itex]f[/itex] is continuous at [itex]y[/itex], we know we can find a [itex]\delta[/itex] such that [itex]|t - y| < \delta[/itex] implies [itex]|f(t) - f(y)| < \epsilon[/itex]. So, if we make [itex]|h| < \delta[/itex], we have
    [tex]
    \begin{align*}
    \left| \frac{1}{h} \int_y^{y+h} f - f(y) \frac{1}{h} \int_y^{y+h} f \right|
    & \leq \frac{1}{h} \int_y^{y+h} \left| f(t) - f(y) \right| dt\\
    & \leq \frac{1}{h} \int_y^{y+h} \epsilon dt = \epsilon.
    \end{align*}
    [/tex]

    That does it, I think.
     
  5. Aug 15, 2010 #4

    hunt_mat

    User Avatar
    Homework Helper

    There is a typo, you should write:
    [tex]
    \left| \frac{1}{h} \int_y^{y+h} f - f(y) \right| =\Bigg|\frac{1}{h}\int_{y}^{y+h}f-f(y)\int_{y}^{y+h}\frac{1}{h}\Bigg|
    [/tex]
     
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