Equivalence of Definitions of Riemann and Darboux Integrals

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a real function.
Then:
:the Riemann integral of $f$ over $\closedint a b$ exists and is equal to $L$
:
:the Darboux integral of $f$ over $\closedint a b$ exists and is equal to $L$.
== Proof ==
=== Riemann Integral $\implies$ Darboux Integral ===
Let $L$ be the Riemann integral of $f$ over $\closedint a b$.
Then $\forall \epsilon > 0: \exists \delta > 0$ such that for every finite subdivision with norm $< \delta$
and every sequence of corresponding sample points, the Riemann sum of the subdivision is in $\openint {L - \epsilon} {L + \epsilon}$.
 the lower Darboux integral of $f$ over $\closedint a b$ is less than $L$.
Let:
:$\ds \epsilon = L - \underline {\int_a^b} \map f x \rd x$
Because $\ds \underline {\int_a^b} \map f x \rd x < L$ it follows that $\epsilon > 0$.
Therefore, there is a $\delta$ that satisfies the conditions above.
Let $n = \ceiling {\dfrac {b - a} \delta}$.
Define:
:$x_i = a + \dfrac i n \paren {b - a}$
Then the sequence $\sequence {x_i}_{0 \mathop \le i \mathop \le n}$ is a finite subdivision of $\closedint a b$.
Additionally, for every $i$:
So the norm of $\sequence {x_i} < \delta$.
Let $K$ be the lower Darboux sum of of $f$ over $\sequence {x_i}$.
The lower Darboux sum is a Riemann sum, so by the definition of $\delta$:
:$\ds K > L - \epsilon = \underline {\int_a^b} \map f x \rd x$
But by the definition of lower Darboux integral:
:$\ds K \le \underline {\int_a^b} \map f x \rd x$
These inequalities are contradictory, so our supposition that the lower Darboux integral is less than $L$ was false.
Therefore:
:$\ds \underline {\int_a^b} \map f x \rd x \ge L$
A similar argument shows that:
:$\ds \overline {\int_a^b} \map f x \rd x \le L$
But by Upper Darboux Integral Never Smaller than Lower Darboux Integral:
:$\ds \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x$
Thus, the following is true:
:$\ds L \le \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x \le L$
and we can conclude that:
:$\ds L = \underline {\int_a^b} \map f x \rd x = \overline {\int_a^b} \map f x \rd x$
=== Darboux Integral $\implies$ Riemann Integral ===
Let $D$ be the Darboux integral of $f$ over $\closedint a b$.
Then for any $\epsilon' > 0$ there is a pair of subdivisions $P$ and $P'$ such that:
:the lower Darboux sum $\map L P > D - \epsilon'$
and:
:the upper Darboux sum $\map U {P'} < D + \epsilon'$.
Fix an $\epsilon > 0$, and choose such subdivisions for $\epsilon' = \dfrac 1 2 \epsilon$.
Let $Q = \sequence {y_0, \dotsc, y_m} = P \cup P'$.
Then from Lower Sum of Refinement and Upper Sum of Refinement, it follows that:
:$D - \dfrac 1 2 \epsilon < \map L Q \le \map U Q < D + \dfrac 1 2 \epsilon$
Let $\delta' = \map \min {y_1 - y_0, \dotsc, y_m - y_{m - 1} }$ be the size of the smallest interval in $Q$.
Let $\ds r = \sup_{\closedint a b} \map f x - \inf_{\closedint a b} \map f x$ be an upper bound for $\size {\map f z - \map f y}$ for all $y, z \in \closedint a b$.
If $r = 0$, we can choose any $r > 0$, as any upper bound will do.
Let $\delta = \map \min {\delta', \dfrac {\epsilon} {2 r \paren {m - 1} } }$.
Fix a subdivision $\Delta = \sequence {x_0, \dotsc, x_n}$ with norm $< \delta$ and sequence of sample points $C = \sequence {t_1, \dotsc, t_n}$, with $x_{i - 1} \le t_i \le x_i$.
Each $\Delta x_i$ is less than every $\Delta y_j$, so for every $i$, either $\closedint {x_{i - 1} } {x_i}$ is contained in some $\closedint {y_{j - 1} } {y_j}$, or $\openint {x_{i - 1} } {x_i}$ contains exactly one point $y_j$.
There are $m - 1$ points that can be contained in an open interval, since $a \le x_i \le b$ for every $i$.
Therefore, there are at most $m - 1$ intervals $\closedint {x_{i - 1} } {x_i}$ which are not contained in a $\closedint {y_{j - 1} } {y_j}$.
Let $\set {i_k}$ be the indices $i$ of those intervals.
Let $j_k$ be the index of the unique $y_{j_k} \in \openint {x_{i_k - 1} } {x_{i_k} }$.
Then, $t_{i_k}$ is in at least one of $\closedint {x_{i_k - 1} } {y_{j_k} }$ and $\closedint {y_{j_k} } {x_{i_k} }$.
Let $\closedint {c_k} {d_k}$ be an interval that contains $t_{i_k}$, choosing the left one if $t_{i_k} = y_{j_k}$, and $\closedint {u_k} {v_k}$ be the other one.
Then:
By the definition of $r$, for any point $s \in \closedint a b$:
Now, it follows that:
where $\paren {c, d} \in \set {\paren {x_{i - 1}, x_i}: i \neq i_k} \cup \set {\paren {c_k, d_k} } \cup \set {\paren {u_k, v_k} }$ and each $t'_i \in \closedint c d$.
Since each $t'_i \in \closedint c d \subset \closedint {y_{j-1} } {y_j}$, it holds that:
:$\ds \inf_{\closedint {y_{j-1} } {y_j} } \map f y \le \map f {t'_i} \le \sup_{\closedint {y_{j - 1} } {y_j} } \map f y$
Thus, $\ds \paren {d - c} \inf_{\closedint {y_{j-1} } {y_j} } \map f y \le \paren {d - c} \map f {t'_i} \le \paren {d - c} \sup_{\closedint {y_{j-1} } {y_j} } \map f y$.
But the intervals $\closedint c d$ cover $\closedint a b$, so:
:$\ds \map L Q = \sum_j \paren {d - c} \inf_{\closedint {y_{j-1} } {y_j} } \map f y \le \sum_{\paren{c,d} } \map f {t'_i} \paren {d - c} \le \sum_j \paren {d - c} \sup_{\closedint {y_{j - 1} } {y_j} } \map f y = \map U Q$
So:
:$\ds \size {D - \sum_{\paren {c, d} } \map f {t'_i} \paren {d - c} } < \frac \epsilon 2$
Therefore:
In other words:
:$\size {\map S {f; \Delta, C} - D} < \epsilon$
where $S$ denotes the Riemann sum.
Because the subdivision and sample points were arbitrary, this holds for every such subdivision and samples.
Because $\epsilon$ was also arbitrary, there is a $\delta$ for every $\epsilon$ such that, if the norm is less than $\delta$, then the Riemann sum is within $\epsilon$ of $D$.
Therefore, $D$ is the Riemann integral of $f$ over $\closedint a b$.
== Sources ==
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Category:Integral Calculus