# Prove that for any $\epsilon > 0, \exists \delta > 0,$ if $||P|| < \delta$, then $|L(f,P) - I|<\epsilon$ , and $|U(f,P) - I|<\epsilon$

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?

• 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
• It is a classic theorem, but the proof is lengthy. Perhaps you should consult a textbook. – Paul Frost Nov 29 '18 at 11:58
• 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 3 '18 at 0:23
• You may want to have a look at the following answer: math.stackexchange.com/a/2047959/72031 – Paramanand Singh Dec 3 '18 at 0:56

## 1 Answer

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 $$|U(f,Q) - U(f,P)| < 2M(m-1) \delta$$.