# 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 $\openint {L - \epsilon} {L + \epsilon}$.

Aiming for a contradiction, suppose the Lower 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 $\sequence {x_i}_{0 \mathop \leq i \mathop \leq n}$ is a Finite Subdivision of $\closedint a b$.

Additionally, for every $i$:

 $\ds x_i - x_{i - 1}$ $=$ $\ds \paren {a + \frac{i}{n} \paren{b - a} } - \paren {a + \frac{i - 1}{n} \paren{b - a} }$ $\ds$ $=$ $\ds \frac{b - a}{n}$ $\ds$ $<$ $\ds \paren{b - a} \frac{\delta}{b - a}$ $\ds$ $=$ $\ds \delta$

So the norm of $\sequence {x_i} < \delta$.

Let $K$ be the Lower Sum of of $f$ over $\sequence {x_i}$.

The Lower 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 Integral:

$\ds K \leq \underline {\int_a^b} \map f x \rd x$

These inequalities are contradictory, so our supposition that the Lower Integral is less than $L$ was false.

Therefore:

$\ds \underline {\int_a^b} \map f x \rd x \geq L$

A similar argument shows that:

$\ds \overline {\int_a^b} \map f x \rd x \leq L$
$\ds \underline {\int_a^b} \map f x \rd x \leq \overline {\int_a^b} \map f x \rd x$

Thus, the following is true:

$\ds L \leq \underline {\int_a^b} \map f x \rd x \leq \overline {\int_a^b} \map f x \rd x \leq L$

and we can conclude that:

$\ds L = \underline {\int_a^b} \map f x \rd x = \overline {\int_a^b} \map f x \rd x$

$\Box$

### 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 Sum $\map L P > D - \epsilon'$ and the Upper 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:

$D - \frac 1 2 \epsilon < \map L Q \leq \map U Q < D + \frac 1 2 \epsilon$

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 $\ds \delta = \map \min {\delta', \frac{\epsilon}{2 r \paren{m - 1}} }$.

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 $x_{i - 1} \leq t_i \leq x_i$.

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 $a \leq x_i \leq b$ 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:

 $\ds \paren {x_{i_k} - x_{i_k - 1} } \map f {t_{i_k} }$ $=$ $\ds \paren {y_{j_k} - x_{i_k - 1} } \map f {t_{i_k} } + \paren {x_{i_k} - y_{j_k} } \map f {t_{i_k} }$ $\ds$ $=$ $\ds \paren {d_k - c_k} \map f {t_{i_k} } + \paren {v_k - u_k} \map f {t_{i_k} }$

By the definition of $r$, for any point $s \in \closedint a b$:

 $\ds \size {\map f {t_{i_k} } - \map f s }$ $\leq$ $\ds r$ Definition of $r$ $\ds \paren {v_k - u_k} \size {\map f {t_{i_k} } - \map f s }$ $\leq$ $\ds \paren {v_k - u_k} r$ $\ds$ $<$ $\ds \paren {x_{i_k} - x_{i_k - 1} } r$ $\closedint {u_k} {v_k} \subsetneq \closedint {x_{i_k - 1} } {x_{i_k} }$ $\ds$ $<$ $\ds \delta r$ Definition of $\sequence {x_i}$ $\ds$ $\leq$ $\ds \frac{\epsilon}{2\paren{m - 1} }$ Definition of $\delta$

Now, it follows that:

 $\ds \sum_{i \mathop = 1}^n \map f {t_i} \paren {x_i - x_{i - 1} }$ $=$ $\ds \sum_{i \mathop \neq i_k} \map f {t_i} \paren {x_i - x_{i - 1} } + \sum_k \map f {t_{i_k} } \paren {x_{i_k} - x_{i_k - 1} }$ $\ds$ $=$ $\ds \sum_{i \mathop \neq i_k} \map f {t_i} \paren {x_i - x_{i - 1} } + \sum_k \paren {\paren {d_k - c_k} \map f {t_{i_k} } + \paren {v_k - u_k} \map f {t_{i_k} } }$ $\ds$ $=$ $\ds \sum_{i \mathop \neq i_k} \map f {t_i} \paren {x_i - x_{i - 1} } + \sum_k \paren {d_k - c_k} \map f {t_{i_k} }$ $\ds$  $\, \ds + \,$ $\ds \sum_k \paren {v_k - u_k} \map f {y_{j_k} } + \sum_k \paren {v_k - u_k} \paren {\map f {t_{i_k} } - \map f {y_{j_k} } }$ $\ds$ $=$ $\ds \sum_{\paren{c,d} } \map f {t'_i} \paren {d - c} + \sum_k \paren {v_k - u_k} \paren {\map f {t_{i_k} } - \map f {y_{j_k} } }$

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 $\ds \inf_{\closedint {y_{j-1} } {y_j} } \map f y \leq \map f {t'_i} \leq \sup_{\closedint {y_{j-1} } {y_j} } \map f y$.

Thus, $\ds \paren {d - c} \inf_{\closedint {y_{j-1} } {y_j} } \map f y \leq \paren {d - c} \map f {t'_i} \leq \paren {d - c} \sup_{\closedint {y_{j-1} } {y_j} } \map f y$.

But the intervals $\closedint c d$ cover $\closedint a b$, so:

$\ds \map L Q = \sum_j \paren {d - c} \inf_{\closedint {y_{j-1} } {y_j} } \map f y \leq \sum_{\paren{c,d} } \map f {t'_i} \paren {d - c} \leq \sum_j \paren {d - c} \sup_{\closedint {y_{j-1} } {y_j} } \map f y = \map U Q$

So:

$\ds \size {D - \sum_{\paren{c,d} } \map f {t'_i} \paren {d - c} } < \frac{\epsilon}{2}$

Therefore:

 $\ds \size {D - \sum_{i \mathop = 1}^n \map f {t_i} \paren {x_i - x_{i - 1} } }$ $\leq$ $\ds \size {D - \sum_{\paren{c,d} } \map f {t'_i} \paren {d - c} } + \size {\sum_k \paren {v_k - u_k} \paren {\map f {t_{i_k} } - \map f {y_{j_k} } } }$ Triangle Inequality $\ds$ $\leq$ $\ds \size {D - \sum_{\paren{c,d} } \map f {t'_i} \paren {d - c} } + \sum_k \paren {v_k - u_k} \size {\map f {t_{i_k} } - \map f {y_{j_k} } }$ Triangle Inequality $\ds$ $<$ $\ds \frac{\epsilon}{2} + \sum_k \frac{\epsilon}{2\paren{m - 1} }$ Identities from above $\ds$ $\leq$ $\ds \frac{\epsilon}{2} + \frac{\epsilon}{2}$ There are at most $m - 1$ values of $k$ $\ds$ $=$ $\ds \epsilon$

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 $\closedint a b$.

$\blacksquare$