How does he do the algebra? (page 134 Rudin, chapter 6 ,theorem 6.20)

$\left| \frac {F(t)- F(s)}{t-s} -f(x_o) \right| = \left| \frac{1}{t-s} \int_s^t[f(u) - f(x_o)]du \right|< \epsilon $

also, how does he conclude that $F'(x_o) = f(x_0)$

: here is the theorem(and the proof by rudin)

Let $f \in \Re$ on $[a,b]$. For $ a \leq x \leq b$, put: $F(x) = \int_a^x f(t)dt$, Then $F$ is continuous on $[a,b]$; furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_o$ and $F'(x_0) = f(x_0)$.

(i will omit the proof of continuity of $F$ on $[a,b]$)

Suppose $f$ is continuous at $x_0$. Given $\epsilon > 0 $ choose $\delta > 0$ such that:

$\vert f(t)- f(x_o) \vert < \epsilon $

if $\vert t- x_0 \vert < \delta$, and $a \leq t \leq b $.Hence, if

$x_0 - \delta < s \leq x_0 \leq t < x_0 + \delta$ $\enspace$ with: $a\ \leq s < t \leq b$

we have by theorem 6.12(d)

$\left| \frac{F(t) - F(s)}{t-s} - f(x_0) \right| = \left| \frac{1}{t-s} \int_s^t [f(u) - f(x_0)]du \right| < \epsilon$

it follows that $F'(x_0) = f(x_0)$

  • $\begingroup$ Could you give some more information, what are you trying to prove, the context etc. $\endgroup$ – user112167 Dec 4 '13 at 23:50
  • 1
    $\begingroup$ For those of us who don't have the book, how is $F$ defined? And how does $x_0$ relate to $t,s$? $\endgroup$ – Daniel Fischer Dec 4 '13 at 23:51
  • $\begingroup$ sorry guys i will write down the hole proof tomorrow so that those that doesnt have the book in hand will get the context , sorry $\endgroup$ – Danny Dec 5 '13 at 0:13
  • $\begingroup$ I have a doubt in the continuity of F. By definition of F we get F(a)=0 which need not be true in general I think. For f(x)=x on [1,2], F(1)=1/2 not= 0. Somebody please explain. $\endgroup$ – supremum Aug 13 '17 at 7:55

You forgot to subtract $f(x_0)$ in the absolute value. First, note that $$F(t)-F(s)=\int_a^tf(u)du-\int_a^sf(u)du=\int_s^tf(u)du. $$ Also, note that since $f(x_0)$ is a constant $$f(x_0)=\frac{1}{t-s} \int_s^t f(x_0)du .$$ Combining the two we find $$\left| \frac{F(t)-F(s)}{t-s}-f(x_0) \right|=\left|\frac{1}{t-s} \int_s^t [f(u)-f(x_0)]du \right|=\frac{1}{|t-s|} \left| \int_s^t [f(u)-f(x_0)]du \right|. $$ The final step is using the "triangle inequality for integrals" (Theorem 6.12(d)); you need to remember that $\left|f(u)-f(x_0)\right| < \varepsilon$ .


I'm not exactly sure of the context, but I can provide some information that I believe is relevant. There appears to be an error in what you have written.

Let $f\in L^{1}(-\infty, \infty)$. Define the function $F(t)=\int_{-\infty}^{t}f(y)dy$. Now let $x_{0}\in [s,t]$ \begin{eqnarray*} \left| \frac{F(t)-F(s)}{t-s}-f(x_{0})\right| &=& \left| \frac{1}{t-s} \int_{s}^{t}f(y)dy-f(x_{0})\right| \\ &=& \left| \frac{1}{t-s}\int_{s}^{t}\left(f(y)-f(x_{0})\right)dy\right| \\ &\leq & \frac{1}{t-s}\int_{s}^{t}\left|f(y)-f(x_{0})\right|dy \end{eqnarray*}

Now if $x_{0}$ is a Lebesgue point, the right hand side converges to $0$ as the length of the interval goes to $0$. More explicitly, for every $\epsilon>0$, there is $\delta$ such that if $t-s<\delta$, the right hand side is $<\epsilon$. Now, by Lebesgue's theorem, almost every point in $(-\infty,\infty)$ is a Lebesgue point. Therefore, the limit on the left hand side converges to $f(x_{0})$ for almost all $x_{0}\in(-\infty,\infty)$. By definition, this limit is $F^{\prime}(x_{0})$.

This is really a proof that the Radon-Nikodym derivative with respect to the Lebesgue measure agrees with the usual definition of derivative as the limit of the difference quotient.

  • $\begingroup$ oops...looks ike user 1337 already gave the result. $\endgroup$ – user113529 Dec 5 '13 at 0:42

about the last step , you mean that:

$\left| \frac{1}{t-s} \int_s^t f(u)-f(x_o)du \right| \leq \frac{1}{t-s} \int_s^t \left| f(u)-f(x_o)du \right| < \frac{1}{t-s} \int_s^t \epsilon = \frac{1}{t-s} \epsilon(t-s)= \epsilon$

  • $\begingroup$ Precisely. Next time try to make it as a comment please. $\endgroup$ – user1337 Dec 5 '13 at 12:16

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