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Let function f be integrable on [a,b] and $I = \int_{a}^{b} f(x) dx.$ Then, for any $\epsilon > 0, \exists \delta > 0,$ such that if P is any partition of [a,b] and $||P|| < \delta $, then $|L(f,P) - I|<\epsilon $ , and $|U(f,P) - I|<\epsilon $

Could anyone give me a hint for this proof?

Intuition
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  • Your notation is not standard so could specify your symbols? – Federico Fallucca Nov 29 '18 at 11:15
  • What is for you the definition of integral of a function? – Federico Fallucca Nov 29 '18 at 11:16
  • Upper darboux sum and Lower one. @FedericoFallucca – Intuition Nov 29 '18 at 11:16
  • integrable in the question means darboux integrable @PaulFrost – Intuition Nov 29 '18 at 11:18
  • @FedericoFallucca integrable in the question means darboux integrable – Intuition Nov 29 '18 at 11:19
  • I think that you must simply use the definition of the sup and inf on the lower sum and upper sum respectively – Federico Fallucca Nov 29 '18 at 11:23
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    It is a classic theorem, but the proof is lengthy. Perhaps you should consult a textbook. – Paul Frost Nov 29 '18 at 11:58
  • could you recommend a book please?@PaulFrost – Intuition Nov 29 '18 at 12:01
  • You can extract it for example from http://math.louisville.edu/~lee/ira/IntroRealAnal-ch08.pdf. The Riemann integral is defined in Definition 8.1, and in Theorem 8.9 it is proved that it agrees with the Darboux integral. Looking at the proof you will see that for $\lVert P \rVert < \delta$, $\delta$ sufficiently small, you get what is desired. – Paul Frost Dec 01 '18 at 17:21
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    This is an aspect of parallel (and equivalent) approaches to the Riemann integral -- convergence of sums as partitions are refined (Darboux) and convergence as partition norm tends to $0$. Most books either fail to mention the connection or if mentioned it is not explained clearly or left as an exercise. The link provided in a comment is somewhat difficult to wade through. My answer shows that there is one key step that is not immediately obvious, but other than that the proof is straightforward. – RRL Dec 03 '18 at 00:23
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    You may want to have a look at the following answer: https://math.stackexchange.com/a/2047959/72031 – Paramanand Singh Dec 03 '18 at 00:56

1 Answers1

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Start with the assumption that $f$ is Riemann-Darboux integrable and, hence, bounded.

For any $\epsilon > 0$ there exists a partition $P_\epsilon = (a=x_0,x_1, \ldots, x_{m-1},x_m=b)$ such that the upper Darboux sum satisfies

$$I \leqslant U(f,P_\epsilon) < I + \frac{\epsilon}{2}$$

Since $f$ must be bounded, there exists $M > 0$ such that $-M \leqslant f(x) \leqslant M$ and $|f(x)- f(y)| \leqslant 2M$ for all $x,y \in [a,b]$.

Let $P = (a = y_0 , y_1, \ldots , y_{r-1}, y_r = b)$ be any partition with $\|P\| < \delta = \dfrac{\epsilon}{4mM},$ and take $Q = P \cup P_\epsilon$.

Since the partition $Q$ is a refinement of $P_\epsilon$ we have $U(f,Q) \leqslant U(f,P_\epsilon)$. Furthermore, $Q$ has at most $m-1$ more partition points than $P$ since the $m+1$ points of $P_\epsilon$ have been added and the endpoints $x_0 = y_0 =a$ and $x_m = y_r = b$ coincide.

The part of the proof that requires some insight is the observation that

$$\tag{*}|U(f,P) - U(f,Q)| < 2M(m-1) \delta = 2M(m-1) \frac{\epsilon}{4mM} < \frac{\epsilon}{2},$$

which implies

$$U(f,P) < U(f,Q) + \frac{\epsilon}{2} < U(f,P_\epsilon) + \frac{\epsilon}{2} < I + \epsilon$$ Since $U(f,P) \geqslant I$ it follows that $|U(f,P) - I| < \epsilon$. The proof that $|L(f,P) - I| < \epsilon$ is similar.

Explanation of inequality (*)

This follows because the difference between $U(f,P)$ and $U(f,Q)$ comes from the area of at most $m-1$ rectangles above the graph of $f$ with height bounded by $2M$ and width bounded by $\delta$.

For example, consider the interval $[y_j, y_{j+1}]$ of $P$ and suppose that the single point $x_k$ from $P_\epsilon$ has been added in forming $Q$ and we have $y_j < x_k < y_{j+1}$.

Let $M(\alpha,\beta) := \sup_{x \in [\alpha,\beta]}\,f(x)$

The absolute difference of upper sums has the contribution

$$|U(f,Q) - U(f,P)| = \left| \,M(y_j,x_k) (x_k - y_j)+ M(x_k,y_{j+1}) (y_{j+1} - x_k) - M(y_j,y_{j+1}) (y_{j+1} - y_j)\, \right| \\ \leqslant |M(y_j,x_k)- M(y_j,y_{j+1})| (x_k - y_j)+ |M(x_k,y_{j+1})- M(y_j,y_{j+1}) |(y_{j+1} - x_k) \\ < |M(y_j,x_k)- M(y_j,y_{j+1})|\delta + |M(x_k,y_{j+1})- M(y_j,y_{j+1})| \delta $$

Of the two terms on the RHS one must vanish where suprema coincide and in the remaining term the difference of suprema is bounded by $2M$.

Thus, $|U(f,Q) - U(f,P)| < 2M \delta$ and proceeding inductively as $m-1$ points are added we have $|U(f,Q) - U(f,P)| < 2M(m-1) \delta$.

RRL
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  • I hope it's okay I ask a question on this old post of yours (+1). What exactly do you mean by proceeding inductively? I see how you showed the base case, $m=1$. But how do you go about showing the inductive step? – psie Apr 26 '23 at 21:03
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    @schn: It just means I am estimating the absolute difference of the upper sum $U(f,P)$ and an upper sum corresponding to a refinement of $P$ obtained by adding $m-1$ points from $P_\epsilon$ in succession. By my argument if I form the partition $Q_1$ that is a refinement of $P$ obtained from adding one point, then $|U(f,Q_1) - U(f,P)| < 2M\delta$. Now add a second point to $Q_1$ to obtain a refinement $Q_2$. By the same argument $|U(f,Q_2)-U(f,Q_1)| < 2M\delta$ which implies $|U(f,Q_2) - U(f,P)| \leqslant |U(f,Q_2) - U(f,Q_1)| + |U(f,Q_1) - U(f,P)| < 2M\delta + 2M\delta = 2M\delta\cdot 2$. – RRL Apr 27 '23 at 05:11
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    Continuing in this way we get the bound $2M\delta \cdot (m-1)$. – RRL Apr 27 '23 at 05:12