3
$\begingroup$

I'm working on proving the following equation:

$\lim_{\delta \to 0^+} \frac{1}{\delta} \int_x^{x + \delta} f(t) \ \mathrm{d}t = f(x)$,

where $f$ is given to be Riemann integrable and continuous on [0,1].

Attempted Proof

I thought about applying the Fundamental Theorem of Calculus to arrive at, for the LHS, something like,

$\lim_{\delta \to 0^+} \frac{F(x + \delta) - F(x)}{\delta}$,

where $F$ is the antiderivative of $f(t).$ However, this feels very circular to me, since this expression of course equals $F'(x) = f(x)$, and we would then have equality on $(0, 1).$

I'd be grateful for any direction. Thanks so much.

$\endgroup$
  • 1
    $\begingroup$ This is not true as written. Consider the function $f(0) = 0, f(x) = 1$ otherwise and look what happens if you evaluate the above expression at $0$. You want $f$ continuous. In that case, think about the mean value theorem. $\endgroup$ – Chris Janjigian Jan 19 '13 at 0:27
  • $\begingroup$ @ChrisJanjigian My apologies, the limit should be as $\delta \to 0^+$, not $x \to 0^+$. Thanks for the catch. $\endgroup$ – James Evans Jan 19 '13 at 0:28
  • $\begingroup$ @dirk5959 Chris is not pointing out anything about $\delta$. Your result is incorrect as Chris has demonstrated it with an example. You need $f$ to be continuous at $x$, to get your conclusion. $\endgroup$ – user17762 Jan 19 '13 at 0:30
  • $\begingroup$ @Marvis Hmm, I am only given integrability in the problem, but I will include continuity in my question since Chris's example does seem sound. $\endgroup$ – James Evans Jan 19 '13 at 0:33
  • $\begingroup$ without continuity this is true for almost every point(lebesgue measure) of $[0,1]$, $\endgroup$ – user52188 Jan 19 '13 at 0:36
4
$\begingroup$

$$ \left|\frac1\delta\int_x^{x+\delta} f(t)\,dt-f(x)\right|\le \sup_{t\in [x,x+\delta]}|f(t)-f(x)|\to 0\mbox{ as $\delta\to 0^+$}$$ where the final limit follows from the continuity of $f$ at $x$.

$\endgroup$
  • $\begingroup$ Many thanks! Is there anyway that the proof could be adapted to show that the statement holds for almost every point of [0,1] if continuity of $f$ is not assumed? $\endgroup$ – James Evans Jan 21 '13 at 18:49
  • $\begingroup$ I'm not sure. But I suspect that more sophisticated tools would be required for that. $\endgroup$ – user108903 Jan 21 '13 at 19:00
1
$\begingroup$

A hint brought to you by nonstandart analysis: you know that $$\int_x^{x+\delta}f(k)\,\mathrm{d}k=F(x+\delta)-F(x),$$ where $F'(x)= f(x)$. Now, use the fact that $$f(a+\varepsilon)\approx f(a)+\varepsilon f'(a)$$ to write $$F(x+\delta)-F(x)\approx F(x)+\delta F'(x)-F(x)=\delta F'(x)=\delta f(x).$$ Try to adapt the use of infinitesimals using limits as $\delta\rightarrow0.$ Of course, this needs some assumptions like continuity and differentiability of $f(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.