User:Caliburn/s/mt/Riemann-Lebesgue Theorem/Proof 2

Proof
From Condition for Darboux Integrability, we have:


 * for all $\epsilon > 0$ there exists a finite subdivision $P$ such that $\map U {f, P} - \map L {f, P} < \epsilon$

where:


 * $\map U {f, P}$ is the upper sum of $f$ with respect to $P$
 * $\map L {f, P}$ is the lower sum of $f$ with respect to $P$.

We now construct a sequence of finite subdivisions $\sequence {P_n}_{n \mathop \in \N}$.

Let $P_1$ be such that:


 * $\map U {f, P_1} - \map L {f, P_1} < 1$

For $n > 1$, let $Q_n$ be such that:


 * $\ds \map U {f, Q_n} - \map L {f, Q_n} < \frac 1 n$

and:


 * $P_n = Q_n \cup P_{n - 1}$

Since $P_n$ is a refinement of $Q_n$, we have:


 * $\map U {f, P_n} \le \map U {f, Q_n}$

and:


 * $\map L {f, P_n} \ge \map L {f, Q_n}$

from Upper Sum of Refinement and Lower Sum of Refinement.

We also have that:


 * $P_{n - 1} \subseteq P_n$

for $n > 1$, so the sequence $\sequence {P_n}_{n \mathop \in \N}$ is increasing.

For each $n$, write:


 * $P_n = \set {a_1, a_2, \ldots, a_{k_n} }$

with:


 * $a = a_1 < a_2 < \ldots < a_{k_n} = b$

For each $1 \le i \le k_n - 1$ define:


 * $m_i = \inf \set {\map f x : x \in \hointl {a_i} {a_{i + 1} } }$

and:


 * $M_i = \set {\map f x : x \in \hointl {a_i} {a_{i + 1} } }$

From Continuous Function on Compact Subspace of Euclidean Space is Bounded, these quantities are finite for each $i$.

For each $n$, define the function $g_n : \closedint a b \to \R$ by:


 * $\ds g_n = \sum_{i \mathop = 1}^{k_n - 1} m_i \chi_{\hointl {a_i} {a_{i + 1} } }$

and the function $h_n : \closedint a b \to \R$ by:


 * $\ds h_n = \sum_{i \mathop = 1}^{k_n - 1} M_i \chi_{\hointl {a_i} {a_{i + 1} } }$

Note that for each $n$ we have:


 * $g_n$ is simple

and:


 * $h_n$ is simple.

So from Simple Function is Measurable, we have:


 * $g_n$ is $\lambda$-measurable

and:


 * $h_n$ is $\lambda$-measurable.

From Integral of Characteristic Function: Corollary, we have:


 * $\ds \int \chi_{\hointl {a_i} {a_{i + 1} } } \rd \lambda = \map \lambda {\hointl {a_i} {a_{i + 1} } } = a_{i + 1} - a_i$

So, from Integral of Integrable Function is Homogeneous, we have:


 * $\ds \int M_i \chi_{\hointl {a_i} {a_{i + 1} } } \rd \lambda = M_i \paren {a_{i + 1} - a_i}$

and:


 * $\ds \int m_i \chi_{\hointl {a_i} {a_{i + 1} } } \rd \lambda = m_i \paren {a_{i + 1} - a_i}$

So, we have:

and: