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

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.

• 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. – Chris Janjigian Jan 19 '13 at 0:27
• @ChrisJanjigian My apologies, the limit should be as $\delta \to 0^+$, not $x \to 0^+$. Thanks for the catch. – James Evans Jan 19 '13 at 0:28
• @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. – user17762 Jan 19 '13 at 0:30
• @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. – James Evans Jan 19 '13 at 0:33
• without continuity this is true for almost every point(lebesgue measure) of $[0,1]$, – user52188 Jan 19 '13 at 0:36

$$\left|\frac1\delta\int_x^{x+\delta} f(t)\,dt-f(x)\right|\le \sup_{t\in [x,x+\delta]}|f(t)-f(x)|\to 0\mbox{ as \delta\to 0^+}$$ where the final limit follows from the continuity of $f$ at $x$.
• 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? – James Evans Jan 21 '13 at 18:49
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 $f(x)$.