User:CircuitCraft/Sandbox

This page is a sandbox for User:CircuitCraft. It will mostly contain incomplete proofs that I'm working on, but want to make public the current state of it.

= 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 $L$.

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 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 $\ds n = \ceiling {\frac{b - a}{\delta} }$.

Define:
 * $\ds x_i = a + \frac{i}{n} \paren{b - a}$

Then the sequence $\sequence {x_i}_{0 \mathop \leq i \mathop \leq n}$ is a Finite Subdivision of $\closedint a b$.

Additionally, for every $i$:

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

But:
 * $\ds \underline {\int_a^b} \map f x \rd x \leq \overline {\int_a^b} \map f x \rd x$

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

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 $\ds \epsilon' = \frac{1}{2}\epsilon$.

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:
 * $\ds D - \frac{1}{2}\epsilon < \map L Q \leq \map U Q < K + \frac{1}{2}\epsilon$

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 $\sequence {x_0, \dotsc, x_n}$ with norm $< \delta$ and sequence of sample points $\sequence {t_1, \dotsc, t_n}$, 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 $\closedint {y_{j - 1} } {y_j}$.

For each $\closedint {x_{i_k - 1} } {x_{i_k} }$ of those intervals, let $y_{j_k}$ be the point it contains.

Then:
 * $\paren {x_{i_k} - x_{i_k - 1} } \map f {t_{i_k} } = \paren {y_{j_k} - x_{i_k - 1} } \map f {t_{i_k} } + \paren {x_{i_k} - y_{j_k} } \map f {t_{i_k} }$

$t_{i_k}$ is in at least one of $\closedint {x_{i - 1} } {y_{j_k} }$ and $\closedint {y_{j_k} } {x_{i_k} }$.

Let $\closedint {c_k} {d_k}$ be that interval, choosing the latter if both hold. Let $\closedint {u_k} {v_k}$ be the other interval.

Then $\paren {x_{i_k} - x_{i_k - 1} } \map f {t_{i_k} } = \paren {d_k - c_k} \map f {t_{i_k} } + \paren {v_k - u_k} \map f {t_{i_k} }$.

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

We can extend this inequality to bound the sum over all values of $k$:

Consider some interval $\closedint {y_{j - 1} } {y_j}$, along with any $\sequence {z_0, \dotsc, z_p }$ and sample points $\sequence {u_1, \dotsc, u_p }$, such that:
 * $y_{j - 1} \leq z_0 < \dotsb < z_p \leq y_j$
 * $z_q - z_{q - 1} < \delta$ for all $1 \leq q \leq p$
 * $z_{q - 1} \leq u_q \leq z_q$ for all $1 \leq q \leq p$

The following inequalities hold: