In this post, we will continue to use the notation $M_i = \sup{f}_{x \in [x_{i-1}, x_i]}$ and $m_i = \inf{f}_{x \in [x_{i-1}, x_i]}$, although this notation will not be used in the entirety of this post, as later on in the post, when we prove 'Darboux integrability implies Riemann integrability', clearly identifying over which interval we are taking the supremum of $f$ will become necessary.
We start by presenting the theorem we will attempt to prove.
Theorem 1.1 - A bounded function $f$ is Riemann Integrable on $[a,b]$ if and only if $f$ is Darboux Integrable on $[a,b]$.
Proof (Theorem 1.1) - We will start by proving that if $f$ is Riemann Integrable on $[a,b]$, it is Darboux Integrable on [a,b].
If $f$ is Riemann Integrable, then as per definition $\displaystyle \lim_{||P|| -> 0} \sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) $ exists.
Let $P = \{x_0 = a, x_1, x_2, ... , x_{n-1}, x_n=b\}$ be a partition of $[a,b]$.
Now, consider $\displaystyle \sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) - \sum_{i=1}^n (M_i)(x_i - x_{i-1}) = \sum_{i=1}^n (f(\xi_i) - M_i)(x_i - x_{i-1})$
Let us choose $\xi_i$ so that $|f(\xi_i) - M_i| < \frac{\epsilon}{b-a}$.
(By Lemma 1.0 this is always possible)
Then we have $ \displaystyle |\sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) - \sum_{i=1}^n (M_i)(x_i - x_{i-1}) | =
| \sum_{i=1}^n (f(\xi_i) - M_i)(x_i - x_{i-1})|
< \sum_{i=1}^n |f(\xi_i) - M_i| |x_i - x_{i-1}|$
(This follows from the triangle inequality)
But we have chosen $\xi_i$ so that $|f(\xi_i) - M_i| < \frac{\epsilon}{b-a}$ for each $i$.
Therefore, $ \displaystyle | \sum_{i=1}^n (f(\xi_i) - M_i)(x_i - x_{i-1})| < \sum_{i=1}^n |f(\xi_i) - M_i| |x_i - x_{i-1}| < ( \frac{\epsilon}{b-a})(\sum_{i=1}^n (x_i - x_{i-1}) ) = \epsilon $
Now recall that by definition, we know that $\lim_{||P|| -> 0} \sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) = \int_{a}^b f $ (as $f$ is Riemann integrable on $[a,b]$)
We will show that $\int_{a}^b f $ is a lower bound of $ U(f,P)$ for any partition $P$ .
In order to do so, we will proceed by contradiction.
Suppose for some $P$, $\int_{a}^b f - U(f,P) > 0$.
As $\lim_{||P|| -> 0} \sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) = \int_{a}^b f $, for $||P|| < \delta$ we can make $| \sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) - \int_{a}^b f | < \int_{a}^b f - U(f,P)$
Consider a refinement of $P$ , $P'$, where $||P'|| < \delta$.
Then for $P'$, we have $| \sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) - \int_{a}^b f | < \int_{a}^b f - U(f,P)$
This implies that $U(f,P) - \int_{a}^b f < \sum_{i=1}^n f(\xi_i)(x_i -x_{i-1}) - \int_{a}^b f$ , which is true if and only if $ U(f,P) < \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) $
Furthermore, by the definition of $U(f,P')$, we know that $\sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) \leq U(f,P')$
Before we continue this proof, we need to establish a lemma about Darboux upper and lower sums and how they behave upon refinement.
Lemma 1.1 - If $P'$ is a refinement of $P$, then $U(f,P') \leq U(f,P)$ and $ L(f,P') \geq L(f,P)$.
Proof (Lemma 1.1) - Let $P = \{x_0 = a, x_1, x_2, ... , x_{n-1}, x_n=b\}$. Suppose $P'$ contains one more point (The cases where $P'$ contains two more points, three more points, can be covered inductively - this will be left to the reader).
Then $P' = \{x_0 = a, x_1, x_2, ...., x_k, y' , x_{k+1}, ... , x_{n-1}, x_n\}$.
$U(f,P') = U(f,P) + \sup f_{x \in [x_k,y']}(y' - x_k) + \sup f_{x \in [y', x_{k+1}]}(x_{k+1} - y') - \sup f_{x \in [x_k, x_{k+1}]}(x_{k+1} - x_k) $.
We will now show that $\sup f_{x \in [x_k,y']}(y' - x_k) + \sup f_{x \in [y', x_{k+1}]}(x_{k+1} - y') - \sup f_{x \in [x_k, x_{k+1}]}(x_{k+1} - x_k) \leq 0$.
Clearly, $\sup f_{x \in [x_k,y']} \leq \sup f_{x \in [x_k, x_{k+1}]}$ and $\sup f_{x \in [y',x_{k+1}]} \leq \sup f_{x \in [x_k, x_{k+1}]}$ as $[x_k, y'] \subset [x_k, x_{k+1}] \ and \ [y', x_{k+1}] \subset [x_k, x_{k+1}]$
Therefore, $\sup f_{x \in [x_k,y']}(y' - x_k) + \sup f_{x \in [y',x_{k+1}]}(x_{k+1} - y') \leq (\sup f_{x \in [x_k, x_{k+1}]}) ( y' - x_k + x_{k+1} - y') = \sup_f{x \in [x_k, x_{k+1}]} (x_{k+1} - x_k)$
Thus $U(f,P') \leq U(f,P)$
$\square$
The proof is essentially the same for Darboux lower sums.
Now, back to our proof, we had deduced that $ U(f,P) < \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) $.
Now, we also know that $U(f,P') \leq U(f,P)$ and so $U(f,P') < \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) $.
But we also know that by definition, we have $\sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) \leq U(f,P')$
This yields a contradiction, and therefore $\int_{a}^b f $ is a lower bound of $ U(f,P)$ for any partition $P$.
Now, we know that we can also make $| U(f,P) - \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) | < \frac{\epsilon}{2} $ given we choose $\xi_i$ carefully, and by choosing a $P$ with $||P||$ small enough, we can make $| \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) - \int_{a}^b f | < \frac{\epsilon}{2}$.
We know that $|U(f,P) - \int_{a}^b f | =
|U(f,P) - \int_{a}^b f + \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) - \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) | $
and $ |U(f,P) - \int_{a}^b f + \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) - \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) | \leq
|U(f,P) - \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1})|
+ | \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) - \int_{a}^b f |
< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
Thus by Lemma 1.0, it is clear that $inf \{U(f,P) | P \ is \ a \ partition \ of \ [a,b] \} = \int_{a}^b f$
$\square$
We can similarly prove that $sup \{L(f,P) | P \ is \ a \ partition \ of \ [a,b] \} = \int_{a}^b f$ (this will be left as an exercise to the reader)
So we have successfully shown that ' Riemann Integrability implies Darboux Integrability ' .
We must now proceed to show that the converse is also true, namely, ' Darboux Integrability implies Riemann Integrability '.
In order to do this, we will establish another lemma.
Lemma 1.2 : Let $P = \{x_0 = a, x_1, .... , x_{n-1}, x_n = b\}$ be a partition of $[a,b]$.
Define $|P| = min \{(x_i - x_{i-1}) | \forall i \in \{1,2,...,n\} \}$ that is $|P|$ is the minimum width of partition $P$.
Suppose $P_0 = \{y_0 = a, y_1, ... , y_{m-1}, y_m = b\}$ is a partition of $[a,b]$ such that $|| P_0 || < |P|$ , that is the maximum width of partition $P_0$ is less than the minimum partition width of $P$.
Then for all $i \in \{1,2,....,n-1\}$ there exists a unique $k \in \{1,2,...,m-1\}$ such that
$y_k \leq x_i$ and $x_i < y_{k+1}$. Furthermore there are no points in $P$ that are between $y_k$ and $x_i$ or between $x_i$ and $y_{k+1}$.
More specifically, there do not exist any $x_j$ such that $x_j \in (y_k, x_i)$ or $x_j \in (x_i, y_{k+1})$.
That is, visually speaking, for every $x_i$ there is one and only one point in $P_0$ , $y_k$, such that $y_k$ is immediately to the left of' $x_i$, and the point to the right of $y_k$ in the partition $P_0$, $y_{k+1}$, is immediately to the right of $x_i$.
Proof (Lemma 1.2) -
Consider $X_i = \{ k \in \{1,2,...,m-1\} | y_k \leq x_i \}$ . $X_i$ is nonempty, because we have $||P_0|| < |P|$, therefore, $y_1 < x_1 \leq x_i \forall i$.
$X_i$ is also a finite subset of $\mathbb {N}$. Therefore there exists a $k_0 \in X_i$ such that $k_0$ is the maximum element of $X_i$. That is, $k_0 = max (X_i) $.
We claim that $y_{k_0 + 1} > x_i $. This claim can be proved by contradiction. Suppose $y_{k_0 + 1} \leq x_i$. But then $k_0 + 1 \in X_i$, this is a contradiction, as we said $k_0 = max (X_i)$.
Therefore, $y_{k_0 + 1} > x_i$. Furthermore, we can also prove that $x_{i-1}$ is not in $(y_{k_0}, x_i)$ and that $x_{i+1}$ is not in $(x_i, y_{k_0 + 1})$ by contradiction.
Suppose, $x_{i-1} \in (y_{k_0}, x_i)$ then $y_{k_0} < x_{i-1}< x_i < y_{k+1}$ and so $x_i - x_{i-1} < y_{k_0+1} - y_{k_0}$. But $||P_0|| < |P|$ . Therefore this yields a contradiction .
Similarly, we can prove that $x_{i+1}$ is not in $(x_i, y_{k_0 + 1})$.
Now, all that is left is to prove the uniqueness of $y_{k_0}$. This can also be done by contradiction.
Suppose there was another point $y_j$ in $P_0$ where $j \neq k_0$ such that $y_j \leq x_i$ and $y_{j+1} > x_i$.
If $j > k_0$ then $y_j > y_{k_0}$, and $y_j \leq x_i$. But we know that $k_0 = max{X_i}$. Therefore this yields a contradiction.
If $j< k_0$ then $y_{j+1} \leq y_k \leq x_k$, so $y_{j+1} \ngtr x_i$, this also yields a contradiction.
So $j = k_0$. But we assumed $j \neq k_0$, yielding another contradiction.
Therefore there does not exist another point $y_j$ in $P_0$ where $j \neq k_0$ such that $y_j \leq x_i$ and $y_{j+1} > x_i$.
$\square$
Without further ado, let us prove that ' Darboux Integrability implies Riemann Integrability '.
First we notice $L(f,P) \leq \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) \leq U(f,P)$ for all partitions $P = \{x_0 = a, x_1, x_2, ... , x_{n-1}, x_n = b\}$ of $[a,b]$.
Thus it suffices to show that $\lim_{||P|| -> 0} U(f,P) = \lim_{||P|| -> 0} L(f,P) = \int_{a}^b f$ in order to show that $\lim_{||P|| -> 0} \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1}) = \int_{a}^b f$, as this is a consequence of the squeeze theorem.
As $\inf {U(f,P)}$ exists, by Lemma 1.0 we may choose a partition $P=\{x_0=a, x_1, ... , x_{n-1}, x_n\}$ such that $U(f,P) - \int_{a}^b f < \frac{\epsilon}{2}$.
Now consider a partition $P' = \{y_0 = a, y_1, ... , y_{m-1}, y_m = b\}$ of $[a,b]$ such that $|| P' || < min (| P | , \frac{\frac{\epsilon}{2}}{3nM})$ where $M>0$ such that $|f(x)| < M$ for all $x \in [a,b]$ ,($f$ is bounded), and $n+1$ is the number of points in the partition $P$ described in the previous paragraph.
From Lemma 1.2, we know that since $\displaystyle || P' || < min (| P | , \frac{\frac{\epsilon}{2}}{3nM}) \leq | P |$, the only difference between $P'$ and $P \cup P'$ is that for each unique $y_{i*}$, such that $y_{i*} \leq x_i$ and $y_{i*+1}> x_i$, the sub-interval $[y_{i*},y_{i*+1}]$ does not belong to $P \cup P'$, rather these sub-intervals are replaced by $[y_{i*}, x_i]$ and $[x_i, y_{i*+1}]$ for each $i \in \{1,2,3,...,n-1\}$.
Therefore, $U(f,P') = U(f, P \cup P') - (\sum_{i=1}^n (\sup f_{x \in [y_{i*}, x_i]})(x_i - y_{i*}) + \sum_{i=1}^n (\sup f_{x \in [x_i, y_{i*+1}]})(y_{i*+1}-x_i) ) + \sum_{i=1}^n (\sup f_{x \in [y_{i*}, y_{i*+1}]})(y_{i*+1} - y_{i*}) $
Furthermore, $ | U(f,P') - U(f, P \cup P') | = | \sum_{i=1}^n (\sup f_{x \in [y_{i*}, y_{i*+1}]})(y_{i*+1} - y_{i*}) - (\sum_{i=1}^n (\sup f_{x \in [y_{i*}, x_i]})(x_i - y_{i*}) + \sum_{i=1}^n (\sup f_{x \in [x_i, y_{i*+1}]})(y_{i*+1}-x_i) ) |$
Now,
$| \sum_{i=1}^n (\sup f_{x \in [y_{i*}, y_{i*+1}]})(y_{i*+1} - y_{i*}) - (\sum_{i=1}^n (\sup f_{x \in [y_{i*}, x_i]})(x_i - y_{i*}) + \sum_{i=1}^n (\sup f_{x \in [x_i, y_{i*+1}]})(y_{i*+1}-x_i) ) |$
is less than or equal to
$ \sum_{i=1}^n |(\sup f_{x \in [y_{i*}, y_{i*+1}]})|(y_{i*+1} - y_{i*}) + \sum_{i=1}^n |(\sup f_{x \in [y_{i*}, x_i]})|(x_i - y_{i*}) + \sum_{i=1}^n |(\sup f_{x \in [x_i, y_{i*+1}]})|(y_{i*+1}-x_i) $
by the triangle inequality
As $|f(x)| < M$ for all $ x \in [a,b]$ we may also deduce that
$ \sum_{i=1}^n |(\sup f_{x \in [y_{i*}, y_{i*+1}]})|(y_{i*+1} - y_{i*}) + \sum_{i=1}^n |(\sup f_{x \in [y_{i*}, x_i]})|(x_i - y_{i*}) + \sum_{i=1}^n |(\sup f_{x \in [x_i, y_{i*+1}]})|(y_{i*+1}-x_i) $
is less than
$ M(\sum_{i=1}^n (y_{i*+1} - y_{i*}) + \sum_{i=1}^n (x_i - y_{i*}) + \sum_{i=1}^n (y_{i*+1}-x_i) )$
Furthermore, as $\displaystyle \sum_{i=1}^n (y_{i*+1} - y_{i*}) + \sum_{i=1}^n (x_i - y_{i*}) + \sum_{i=1}^n (y_{i*+1}-x_i) < 3(\sum_{i=1}^n ||P'||) < 3(\sum_{i=1}^n \frac{\frac{\epsilon}{2}}{3nM}) = \frac{\frac{\epsilon}{2}}{M}$
It follows that, $ \displaystyle M(\sum_{i=1}^n (y_{i*+1} - y_{i*}) + \sum_{i=1}^n (x_i - y_{i*}) + \sum_{i=1}^n (y_{i*+1}-x_i) ) < \frac{\epsilon}{2}$.
So we have shown that $| U(f,P') - U(f, P \cup P') | < \frac{\epsilon}{2}$.
Now, we know that $|U(f,P') - \int_{a}^b f | = |U(f,P') - \int_{a}^b f + U(f,P \cup P') - U(f, P \cup P')|$
Now, by the triangle inequality, $|U(f,P') - \int_{a}^b f + U(f,P \cup P') - U(f, P \cup P')| < |U(f,P') - U(f, P \cup P')| + |U(f,P \cup P') - \int_{a}^b f|$.
$P \cup P'$ is a refinement of $P$ , as $ P \subset P \cup P'$.
Therefore, by Lemma 1.1, we know that $|U(f,P \cup P') - \int_{a}^b f| < \frac{\epsilon}{2}$, as we choose $P$ such that $|U(f,P) - \int_{a}^b| < \frac{\epsilon}{2}$.
So, $|U(f,P') - U(f, P \cup P')| + |U(f,P \cup P') - \int_{a}^b f| < \frac{\epsilon}{2} + \frac{\epsilon}{2}$.
Thus, for an arbitrary partition $P'$ such that $ \displaystyle ||P'|| < min (| P | , \frac{\frac{\epsilon}{2}}{3nM})$, we know that
$|U(f,P') - \int_{a}^b f | < |U(f,P') - U(f, P \cup P')| + |U(f,P \cup P') - \int_{a}^b f| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.
Thus, by definition, $\lim_{||P|| - > 0} U(f,P) = \int_{a}^b f$.
The proof that $\lim_{||P|| - > 0} L(f,P) = \int_{a}^b f$ uses exactly the same method.
Therefore, $\lim_{||P|| -> 0} U(f,P) = \lim_{||P|| -> 0} L(f,P) = \int_{a}^b f$
$\square$
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