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If $f$ is Riemann integrable, prove Upper Riemann Integral is equal to Lower Riemann Integral, $U(f) = L(f)$, where

$$U(f) = \inf \{ U(f, P) \mid P \text{ is a partition of } [a,b] \}$$

$$L(f) = \sup \{ L(f, P) \mid P \text{ is a partition of } [a,b] \}.$$

My professor asks us to prove this in parts.

(a) If $\varepsilon \gt 0 $, show that there is a partition of $P_1$ with width $W(P_1) \lt \delta$ and sample points for $P_1$ such that Riemann Sum $S_1 \gt U(f) -\frac{\varepsilon}{2}$ . (Hint: choose the partition based on denition of greatest lower bound, add points to make the width less than $\delta$)

points to make the width less

(b) show that there is a partition of $P_2$ with width $W(P_2) \lt \delta$ and sample points for $P_2$ such that Riemann Sum $S_2 \lt L(f) +\frac{\varepsilon}{2}$.

And another two parts..

I am having problems with proving both (a) and (b).

Edit: While it seems similar to this question, I am supposed to show that $U(f) = L(f)$ and in the addressed question, I don't think $U(f) = L(f)$ was proven.

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abuchay
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    What you are supposed to show here is known to me as the definition of the Riemann Integral (the Riemann integral being the common value of $\inf$ and $\sup$). Which definition are you using? – Thomas May 18 '16 at 16:01
  • Yeah i know what you meant. The one I think I am supposed to use is "$f$ is Riemann integrable, if there exists $A \in R $ such that for any $\epsilon > 0 $ there exists $\delta > 0 $ such that $|S - A | < \epsilon $ whenever S is a Riemann sum for $f $corresponding to any partition of [a, b] of width less than $\delta$. – abuchay May 18 '16 at 16:05
  • One thing probably useful is that for finer partitions, we have both sharper U and L. – Vim May 18 '16 at 16:36
  • @ Thomas While it seems similar to this question, I am supposed to show that $U(f) = L(f)$ and in the addressed question, I don't think $U(f) = L(f)$ was proven. – abuchay May 18 '16 at 17:00
  • @Tim in the addressed question it was shown that one of $U(f), L(f)$ is equal to the Riemann integral with the additional hint that the same reasoning applies to the other quantity, which is then implying $U=L$. – Thomas May 18 '16 at 18:46

1 Answers1

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(I think that the inequalities in the hints are reversed of what they should be: for isntance, $U(f)$ is already less than all the upper sums)

Fix $\varepsilon>0$. Since $f$ is Riemann integrable, by your definition there exists a number $A$ andn $\delta>0$ such that $|S-A|<\varepsilon/4$ for every Riemann sum form a partition of width less than $\delta$.

By definition of infimum there exists a partition $P_1$ such that $$\tag{1}U(f)+\frac\varepsilon4 > U(f,P_1).$$ By subdividing the partition, the upper sum either stays the same or decreases: if $m=\max\{f(t):\ x_1\leq t\leq x_2\}$ and $m_1=\max\{f(t):\ x_1\leq t\leq x_1'\}$, $m_2=\max\{f(t):\ x_1'\leq t\leq x_2\}$, then $$ m_1(x_1'-x_1)+m_2(x_2-x_1')\leq m(x_1'-x_1)+m(x_2-x_1')=m(x_2-x_1). $$ So we may assume, by subdividing $P_1$, that its width is less then $\delta$. This implies that $$\tag{2} |U(f,P_1)-A|<\frac\varepsilon4. $$

Similarly, by definition of supremum there exists a partition $P_2$ such that $$\tag{3} L(f)-\frac\varepsilon4<L(f,P_2). $$ Again, by subdividing, we may assume that the width of $P_2$ is less than $\delta$ (note that subvididing the partition increases the lower sum, so the inequality $(3)$ is preserved. Again by definition of Riemman integrable, $$\tag{4} |A-L(f,P_2)|<\frac\varepsilon4. $$

Now \begin{align} |U(f)-L(f)|&\leq|U(f)-U(f,P_1)|+|U(f,P_1)-L(f,P_2)|+|L(f,P_2)-L(f)|\\ \ \\ &<\frac\varepsilon4+|U(f,P_1)-L(f,P_2)|+\varepsilon4\\ \ \\ &\leq\frac\varepsilon2+|U(f,P_1-A|+|A-L(f,P_2)|\\ \ \\ &<\frac\varepsilon2+\frac\varepsilon4+\frac\varepsilon4=\varepsilon. \end{align} As $\varepsilon$ was arbitrary, $U(f)=L(f)$.

Martin Argerami
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