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The function $f:ℝ→ℝ$ is continuous on $x_0\inℝ$. Prove using the definition of a Darboux Integral that $$\lim_{h→0}∫_{x_0}^{x_0+h}\frac{f(t)}{h}=f(x_0)$$

I'm a first grade math student following an analysis course. The book that is used is Elementary Analysis by Ross.


  1. The upper Darboux sum $U(f,P)$ of $f$ with respect to a partition $P$ is the sum $$U(f,P)=∑_{k=1}^nM(f,[t_{k-1},t_k])(t_k-t_{k-1})$$

  2. The lower Darboux sum $L(f,P)$ of $f$ with respect to a partition $P$ is the sum $$L(f,P)=∑_{k=1}^nm(f,[t_{k-1},t_k])(t_k-t_{k-1})$$

  3. $f$ is continuous at $x_0$ ⇔ $∀ε>0,∃δ>0,(|x-x_0|<δ ⇒ |f(x)-f(x_0)|<ε)$
  4. Let $f$ be a function defined on on $J-\{a\}$ for some interval $J$ containin $a$, and let $L$ be a real number. Then $ \lim_{x→a}f(x)=L$ if and only if $$∀ε>0,∃δ>0,(0<|x-a|<δ⇒|f(x)-L|<ε)$$

Can someone check if this is an correct proof ?


Let $ε>0$. Then there exist an $δ>0$, such that: $|x-x_0|<δ ⇒ |f(x)-f(x_0)|<ε$. Let $0<|h-0|<δ$. If $x\in[x_0,x_0+h]$ then $x\in(x_0-δ,x_0+δ)$, then $|f(x)-f(x_0)|<ε$, then $|\frac{f(x)}{h}-\frac{f(x_0)}{h}|<\frac{ε}{h}$.

Therefore: \begin{equation*} m ( \frac{f(x)}{h},[x_0,x_0+h]) \cdot (x_0+h - x_0) ≥ \frac{f(x_0)-ε}{h} \cdot h = f(x_0)-ε \end{equation*} \begin{equation*} M ( \frac{f(x)}{h},[x_0,x_0+h]) \cdot (x_0+h - x_0) ≤ \frac{f(x_0)+ε}{h} \cdot h = f(x_0)+ε \end{equation*}

Therefore we can conclude that for a partition $P$ of $[x_o,x_0+h]$

\begin{equation*} f(x_0)-ε< L(f,P)≤∫_{x_0}^{x_0+h}\frac{f(t)}{h}≤U(f,P)<f(x_0)+ε \end{equation*} QED

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@Kasper As you can see I deleted my comment when I noticed that. To the OP, this seems pretty much correct to me – Nameless Dec 16 '12 at 18:27
You might want to explain why $U(f,P)<f(x_0)+\epsilon$ (and also the inequality for the lower sum). But, you should find this easy. – David Mitra Dec 16 '12 at 18:31
@DavidMitra Here are some things I still doubt: 1. Shouldn't I say: \begin{equation*} U(f,P) ≥ m ( \frac{f(x)}{h},[x_0,x_0+h]) \cdot (x_0+h - x_0) > \frac{f(x_0)-ε}{h} \cdot h = f(x_0)-ε \end{equation*} Instead of ≥ and > switched. 2. $x\in[x_0,x_0+h]$ doesn't work when h<0. Shouldn't I say something about that ? – Kasper Dec 16 '12 at 21:49
I mean $L(f,P)$ instead of $U(f,P)$ – Kasper Dec 16 '12 at 22:09
@Kasper Yes, you need to worry about the sign of $h$. Perhaps it's easiest to do the left and right hand limits separately. In your inequalities, for $h>0$, you want to say $L(\color{maroon}{f/h},P)> f(x_0)-\epsilon$ and $U(\color{maroon}{f/h},P)< f(x_0) +\epsilon$. – David Mitra Dec 16 '12 at 22:42
up vote 1 down vote accepted

What you did is correct. Here is a simpler, but not by definition, proof:

Let $F(h)=\int_{x_0}^{x_0+h}f(x)\, dx$. Since $f$ is continuous at $x_0$ by the 1st FTC, $F^{\prime}(0)=f(h)$. Therefore,, $$\lim_{h\to 0}\frac{F(h)}{h}=\lim_{h\to 0}\frac{F(h)-F(0)}{h}=F^{\prime}(0)=f(x_0)$$ Again, this is a not by definition proof.

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