# Proving that a function is Riemann Integrable

The usual definition to the Riemann integral is: for every $ε>0$, there exists $\delta$ such that if $P$ is a partition of $[a,b]$, and $\|P\|<\delta$, then $|S(f;P)-s|<\epsilon$. Then $f$ is Riemann Integrable on $[a,b]$ with integral value $s$. But to prove that a function is Riemann Integrable on $[a,b]$, wouldn't it suffice to show that $(A)$ for every $ε>0$ there exists a unique $s$, and a single partition $Q$ of $[a,b]$ such that $|S(f;Q)-s|<ε$? Here $S(f;Q)$ denotes any Riemann sum of $Q$. My proof is as follows: We have that $(B)$ $f$ is Riemann Integrable (according to the usual definition) on $[a,b]$ if and only if for every $ε>0$ there exists a partition $P$ of $[a,b]$ such that $U(f;P)-L(f;P)<ε$ , where $U(f;P)$ and $L(f;P)$ denotes the upper and lower Darboux sum of $P$ respectively. Also, $(C)$ $U(f;P)$ is the supremum of the set of all Riemann sums of $P$, and $L(f;P)$ is the infimum of the set of all Riemann sums of $P$. By $(A)$, we have $s-ε/2<S(f;Q)<s+ε/2$. Since $S(f;Q)$ is any Riemann sum of $Q$, $s+ε/2$ and $s-ε/2$ is an upperbound and a lower bound of the set of all Riemann sums of $Q$ respectively. By $(C)$, $U(f;Q)$ is the supremum of the set of all Riemann sums $S(f;Q)$, and $L(f;Q)$ is the infimum of the set of all Riemann sums $S(f;Q)$. Hence by the properties of the supremum and infimum , we have that $U(f;Q)≤s+ε/2$ and $L(f;Q)≥s-ε/2$. Hence for every $ε>0$, there exists a partition $Q$ of $[a,b]$ such that $U(f;Q)-L(f;Q)≤(s+ε/2)-(s-ε/2)=ε$. Thus by $(B)$, $f$ is Riemann Integrable according to usual definition. Is my proof correct?

• What does "we have s-ε0" mean? $\;$ – user57159 Jul 27 '15 at 11:53
• I'm sorry, I guess that was a typo. Mr. Michael edited it for me – Luca123 Jul 27 '15 at 11:59
• Note: You are relying on the fact that the first definition of Riemann integrability is the same as the upper/lower sum definition. So you are burying the major details in the proof of that equivalence. – DisintegratingByParts Jul 30 '15 at 8:57

Your definition has problems. The most obvious one is that there is nothing in your definition forcing you to "get essentially the same result even when you refine the partition more". But even if you introduce that, you have a more subtle problem. If you specify a fixed partitioning and point evaluation scheme, then certain functions are integrable that shouldn't be. For instance, the indicator function of the rationals in $[0,1]$ is "integrable" in the sense that the left Riemann sums for uniform partitions are all equal to $1$. But it shouldn't be Riemann integrable (it is "too wildly discontinuous").

Being more precise about these problems, we don't want our definition of the Riemann integral to depend on any particular scheme for partitioning the interval. We also don't want it to depend on any scheme for choosing the points in the subintervals to evaluate the integrand. For example, we should not claim that a function is really integrable merely because its left Riemann sums over uniform partitions converge. To enforce this, the definition involves a quantifier over all possible partitions with small mesh and all possible ways to choose evaluation points for such partitions.

One of the basic theorems that makes Riemann integration manageable says that this definition is redundant. This theorem says that Riemann integration and Darboux integration are equivalent. Darboux integration is based on the simpler requirement that for any tolerance, there exists some partition such that the difference between the lower and upper sums for that partition is smaller than the tolerance. Thus, we are free to pick the partition ourselves, but we are forced to check the "worst possible" scheme of evaluation points, rather than all possible ones.

Like your definition, this definition also ostensibly doesn't force you to "get essentially the same result upon refinement", but actually it does. The basic fact, which is key to the proof, is that the upper sum decreases upon refinement, while the lower sum increases upon refinement. Thus the difference goes down upon refinement. So now to prove this result, you need only prove that your definition of integrability is equivalent to the "refinement" definition, which is given after the third paragraph of https://en.wikipedia.org/wiki/Riemann_integral#Riemann_integral

"(C) $U(f;P)$ is the supremum of the set of all Riemann sums,
and $L(f;P)$ is the infimum of the set of all Riemann sums",
not just the set of Riemann sums for partitions "$P$ such that $|S(f;P)-s|<ε$".

Let $U$ be the supremum of the set of all elements of $[0,\hspace{-0.05 in}4]$ and let $L$ be the infimum of the set of all elements of $[0,\hspace{-0.05 in}4]$. $\;\;\;$ To prove that $\: U\hspace{-0.04 in}-\hspace{-0.04 in}L \leq 2 \:$, $\:$ "wouldn't it suffice to show that (A)" "there exists a single" element $x$ of $[0,\hspace{-0.05 in}4]$ such that $\: |\hspace{.03 in}x\hspace{-0.04 in}-\hspace{-0.04 in}2\hspace{.03 in}| < 1$? $\;\;\;$ Here $x$ denotes any element of $[0,\hspace{-0.05 in}4]$. $\;\;\;$ "My proof is as follows:" "By (A), we have" $\: 2 \hspace{-0.04 in}-\hspace{-0.04 in}1 < x < 2\hspace{-0.04 in}+\hspace{-0.05 in}1 \;$. $\;\;\;$ Recall that $U$ "is the supremum of the set of all" elements of $[0,\hspace{-0.05 in}4]$ and $L$ "is the infimum of the set of all" elements of $[0,\hspace{-0.05 in}4]$. $\;\;\;$ "Hence by the properties of the supremum and infimum , we have that" $\: U \leq 2\hspace{-0.04 in}+\hspace{-0.05 in}1 \:$ and $\: L \geq 2 \hspace{-0.04 in}-\hspace{-0.04 in}1 \;$. $\;\;\;$ Hence $\: U\hspace{-0.04 in}-\hspace{-0.04 in}L \leq (2\hspace{-0.04 in}+\hspace{-0.05 in}1)-(2\hspace{-0.04 in}-\hspace{-0.05 in}1) = 2 \;$.

• I edited my post. – Luca123 Jul 27 '15 at 12:20
• I edited my answer. $\;$ – user57159 Jul 27 '15 at 12:43
• However, $s+ε$ and is the upper bound of the set of all Riemann sums of $Q$, since $S(f;Q)$ is any arbitrary Riemann sum of $Q$. Since $U(f;Q)$ is the supremum of the set of all Riemann sums of $Q$, then $U(f;P)≤s+ε$. – Luca123 Jul 27 '15 at 12:56
• Ah, I now realize that there is an interpretation which would make your proof correct. $\hspace{1.34 in}$ – user57159 Jul 27 '15 at 13:04