Approximation of Lower Darboux Integral by Continuous Function

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Theorem

Let $f : \closedint a b \to \R$ be a bounded real function.

Then, for each $\epsilon > 0$, there is a continuous real function $g : \closedint a b \to \R$ such that:

$\paren 1 \quad$ For all $x \in \closedint a b$, $\map g x \le \map f x$.
$\paren 2 \quad$ The infimums of $f$ and $g$ on $\closedint a b$ are the same.
$\paren 3 \quad \ds \underline {\int_a^b} \map f x \rd x < \int_a^b \map g x \rd x + \epsilon$.

where $\ds \underline {\int_a^b}$ denotes the lower Darboux integral, and $\ds \int_a^b$ the ordinary Darboux integral.


Proof

By definition of lower Darboux integral, there is a subdivision $P$ of $\closedint a b$ such that:

$\ds \underline {\int_a^b} \map f x \rd x < \map L P + \frac \epsilon 2$

where $\map L P$ denotes the lower Darboux sum of $f$ with respect to $P$.


Write $P = \set {x_0, x_1, \dotsc, x_n}$.

For each $\nu \in \set {1, 2, \dotsc, n}$, let $m_\nu$ be the infimum of $f$ on $\closedint {x_{\nu - 1}} {x_\nu}$.

Then:

$\ds \map L P = \sum_{\nu \mathop = 1}^n m_\nu \paren {x_\nu - x_{\nu - 1}}$


For each $\nu \in \set {1, 2, \dotsc, n - 1}$, define:

$\mu_\nu = \map \min{m_\nu, m_{\nu + 1}}$
$\Delta_\nu = \map \min {\dfrac \epsilon {n \paren {\size {m_{\nu + 1} - m_\nu} + 1}}, \dfrac {x_\nu - x_{\nu - 1}} 2, \dfrac {x_{\nu + 1} - x_\nu} 2}$

and additionally define:

$\Delta_0 = \Delta_n = 0$


We can now construct the function $g : \closedint a b \to \R$ in a piecewise manner.

For each $\nu \in \set {1, 2, \dotsc, n}$, we define $g$ for all $x \in \closedint {x_{\nu - 1} + \Delta_{\nu - 1}} {x_\nu - \Delta_\nu}$ as:

$\map g x = m_\nu$

For $\nu \in \set {1, 2, \dotsc, n - 1}$, we define $g$ on $\closedint {x_\nu} {x_\nu + \Delta_\nu}$ as:

$\map g x = \mu_\nu + \dfrac {m_{\nu + 1} - \mu_\nu} {\Delta_\nu} \paren {x - x_\nu}$

and define $g$ on $\closedint {x_\nu - \Delta_\nu} {x_\nu}$ as:

$\map g x = \mu_\nu - \dfrac {m_\nu - \mu_\nu} {\Delta_\nu} \paren {x - x_\nu}$


To see that $\paren 1$ holds, note that on any interval $\closedint {x_{\nu - 1}} {x_\nu}$, we have:

$\map g x \le m_\nu \le \map f x$

It is also clear that $\paren 2$ holds, since the value of $\map g x$ is never less than some $m_\nu$, which itself is an infimum of $f$ on some subset of $\closedint a b$.


It remains to prove $\paren 3$.

Consider the integral:

$\ds \int_a^b \map g x \rd x$

By Sum of Integrals on Adjacent Intervals for Continuous Functions, we will break it into parts of the forms:

$\paren C \quad \ds \int_{x_{\nu - 1} + \Delta_{\nu - 1}}^{x_\nu - \Delta_\nu} \map g x \rd x$

and:

$\paren L \quad \ds \int_{x_\nu - \Delta_\nu}^{x_\nu + \Delta_\nu} \map g x \rd x$

so that we can write:

$\ds \int_a^b \map g x \rd x = \sum_{\nu = 1}^n \int_{x_{\nu - 1} + \Delta_{\nu - 1}}^{x_\nu - \Delta_\nu} \map g x \rd x + \sum_{\nu = 1}^{n - 1} \int_{x_\nu - \Delta_\nu}^{x_\nu + \Delta_\nu} \map g x \rd x$


On each integral $\paren C$, $g$ is constant.

So, by Definite Integral of Constant:

$\ds \int_{x_{\nu - 1} + \Delta_{\nu - 1}}^{x_\nu - \Delta_\nu} \map g x \rd x = \paren {x_\nu - x_{\nu - 1}} m_\nu - \Delta_\nu m_\nu - \Delta_{\nu - 1} m_\nu$


On each integral $\paren L$, consider the two cases $\mu_\nu = m_\nu$ and $\mu_\nu = m_{\nu + 1}$.

If $\mu_\nu = m_\nu$, then:

\(\ds \int_{x_\nu - \Delta_\nu}^{x_\nu + \Delta_\nu} \map g x \rd x\) \(=\) \(\ds \int_{x_\nu - \Delta_\nu}^{x_\nu} \map g x \rd x + \int_{x_\nu}^{x_\nu + \Delta_\nu} \map g x \rd x\) Sum of Integrals on Adjacent Intervals for Continuous Functions
\(\ds \) \(=\) \(\ds \Delta_\nu m_\nu + \int_{x_\nu}^{x_\nu + \Delta_\nu} m_\nu + \frac {m_{\nu + 1} - m_\nu} {\Delta_\nu} \paren {x - x_\nu} \rd x\) Definition of $g$ and Definite Integral of Constant
\(\ds \) \(=\) \(\ds 2 \Delta_\nu m_\nu + \frac {m_{\nu + 1} - m_\nu} {\Delta_\nu} \int_0^{\Delta_\nu} x \rd x\) Linear Combination of Definite Integrals and Change of Limits of Integration
\(\ds \) \(=\) \(\ds 2 \Delta_\nu m_\nu + \frac {m_{\nu + 1} - m_\nu} 2 \Delta_\nu\) Integral of Power
\(\ds \) \(=\) \(\ds \Delta_\nu m_\nu + \Delta_\nu m_{\nu + 1} - \frac {m_{\nu + 1} - m_\nu} 2 \Delta_\nu\)

The case $\mu_\nu = m_{\nu + 1}$ is similar, giving:

\(\ds \int_{x_\nu - \Delta_\nu}^{x_\nu + \Delta_\nu} \map g x \rd x\) \(=\) \(\ds 2 \Delta_\nu m_{\nu + 1} - \frac {m_\nu - m_{\nu + 1} } {\Delta_\nu} \int_{-\Delta_\nu}^0 x \rd x\)
\(\ds \) \(=\) \(\ds 2 \Delta_\nu m_{\nu + 1} + \frac {m_\nu - m_{\nu + 1} } 2 \Delta_\nu\)
\(\ds \) \(=\) \(\ds \Delta_\nu m_\nu + \Delta_\nu m_{\nu + 1} - \frac {m_\nu - m_{\nu + 1} } 2 \Delta_\nu\)

By inspecting which of $m_\nu$ and $m_{\nu + 1}$ is greater, we find that in each case:

$\ds \int_{x_\nu - \Delta_\nu}^{x_\nu + \Delta_\nu} \map g x \rd x = \Delta_\nu m_\nu + \Delta_\nu m_{\nu + 1} - \frac {\size {m_{\nu + 1} - m_\nu} } 2 \Delta_\nu$


Combining the above:

\(\ds \int_a^b \map g x \rd x\) \(=\) \(\ds \sum_{\nu = 1}^n \int_{x_{\nu - 1} + \Delta_{\nu - 1} }^{x_\nu - \Delta_\nu} \map g x \rd x + \sum_{\nu = 1}^{n - 1} \int_{x_\nu - \Delta_\nu}^{x_\nu + \Delta_\nu} \map g x \rd x\)
\(\ds \) \(=\) \(\ds \sum_{\nu = 1}^n \paren {\paren {x_\nu - x_{\nu - 1} } m_\nu - \Delta_\nu m_\nu - \Delta_{\nu - 1} m_\nu} + \sum_{\nu = 1}^{n - 1} \paren {\Delta_\nu m_\nu + \Delta_\nu m_{\nu + 1} - \frac {\size {m_{\nu + 1} - m_\nu} } 2 \Delta_\nu}\)
\(\ds \) \(=\) \(\ds \sum_{\nu = 1}^n \paren {x_\nu - x_{\nu - 1} } m_\nu - \frac 1 2 \sum_{\nu = 1}^{n - 1} \size {m_{\nu + 1} - m_\nu} \Delta_\nu\) Since $\Delta_0 = \Delta_n = 0$
\(\ds \leadsto \ \ \) \(\ds \map L P\) \(=\) \(\ds \int_a^b \map g x \rd x + \frac 1 2 \sum_{\nu = 1}^{n - 1} \size {m_{\nu + 1} - m_\nu} \Delta_\nu\) Definition of $\map L P$
\(\ds \) \(\le\) \(\ds \int_a^b \map g x \rd x + \frac 1 2 \sum_{\nu = 1}^{n - 1} \frac \epsilon n\) Definition of $\Delta_\nu$
\(\ds \) \(<\) \(\ds \int_a^b \map g x \rd x + \frac \epsilon 2\)
\(\ds \leadsto \ \ \) \(\ds \underline {\int_a^b} \map f x \rd x\) \(<\) \(\ds \map L P + \frac \epsilon 2\) Choice of $P$
\(\ds \) \(<\) \(\ds \int_a^b \map g x \rd x + \epsilon\) Above

$\blacksquare$

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