STEM Problem Detail

Problem Statement

Prompt

Provide me step by step solution of PROBLEM 7-19.

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**Solution.**
The required proof may be divided into three parts:
(i) the definition of the upper and lower Stieltjes integrals,
(ii) subsidiary lemmas, and
(iii) the actual proof.

**i)** Let
\[
P = \{ a = x_0, x_1, \ldots, x_n = b \}
\]
be a partition of $$[a,b]$$. Let
\[
\Delta g_i = g(x_i) - g(x_{i-1}),
\]
\[
M_i = \sup \{ f(x) : x \in (x_{i-1}, x_i) \},
\quad
m_i = \inf \{ f(x) : x \in (x_{i-1}, x_i) \}.
\]

Then
\[
L(P,f,g) = \sum_{i=1}^{n} m_i \Delta g_i,
\quad
U(P,f,g) = \sum_{i=1}^{n} M_i \Delta g_i.
\]

The upper and lower Stieltjes integrals of $$f$$ with respect to $$g$$ are defined by
\[
\overline{\int_a^b} f \, dg = \inf U(P,f,g),
\]
\[
\underline{\int_a^b} f \, dg = \sup L(P,f,g).
\]

If these are equal, $$f$$ is Stieltjes integrable with respect to $$g$$ on $$[a,b]$$, and the
Stieltjes integral is denoted by
\[
\int_a^b f \, dg.
\]

**ii)** The integral $$\int_a^b f \, dg$$ exists if and only if, given $$\varepsilon > 0$$,
there exists a partition $$P$$ such that
\[
U(P,f,g) - L(P,f,g) < \varepsilon.
\]

To prove this, note that for every partition $$P$$,
\[
L(P,f,g) \le \underline{\int_a^b} f \, dg
\le \overline{\int_a^b} f \, dg \le U(P,f,g).
\]

In particular, if
\[
U(P,f,g) - L(P,f,g) < \varepsilon,
\]
then
\[
0 \le \overline{\int_a^b} f \, dg - \underline{\int_a^b} f \, dg < \varepsilon.
\]

Since this holds for every $$\varepsilon > 0$$, the two integrals are equal.

Conversely, assume the integral exists. Let $$\varepsilon > 0$$. Then there exist partitions
$$P_1$$ and $$P_2$$ such that
\[
U(P_2,f,g) - \int_a^b f \, dg < \varepsilon/2,
\]
\[
\int_a^b f \, dg - L(P_1,f,g) < \varepsilon/2.
\]

Let $$P^*$$ be the common refinement of $$P_1$$ and $$P_2$$. Then
\[
L(P_1,f,g) \le L(P^*,f,g),
\quad
U(P_2,f,g) \ge U(P^*,f,g).
\]

Hence
\[
U(P^*,f,g) - L(P^*,f,g) < \varepsilon.
\]

**Lemma.**
Suppose the above condition holds for given $$\varepsilon$$ and some partition
$$P = \{x_0,\ldots,x_n\}$$. If $$\xi_i \in (x_{i-1},x_i)$$, then
\[
\sum_{i=1}^{n} | f(\xi_i) - f(t_i) | \, \Delta g_i < \varepsilon.
\]

Since $$f(\xi_i)$$ and $$f(t_i)$$ lie between $$m_i$$ and $$M_i$$,
\[
| f(\xi_i) - f(t_i) | \le M_i - m_i.
\]
Thus,
\[
\sum_{i=1}^{n} | f(\xi_i) - f(t_i) | \, \Delta g_i
\le \sum_{i=1}^{n} (M_i - m_i)\Delta g_i
= U(P,f,g) - L(P,f,g) < \varepsilon.
\]

**iii)** Let $$\varepsilon > 0$$. By the lemma, there exists a partition $$P$$ such that
\[
U(P,g') - L(P,g') < \varepsilon.
\]

By the mean value theorem, there exist points $$t_i \in (x_{i-1},x_i)$$ such that
\[
\Delta g_i = g'(t_i)\Delta x_i.
\]

Let $$\xi_i \in (x_{i-1},x_i)$$. Then
\[
\sum_{i=1}^{n} f(\xi_i)\Delta g_i
= \sum_{i=1}^{n} f(\xi_i) g'(t_i)\Delta x_i.
\]

Let $$M = \sup |f(x)|$$. Then
\[
\left|
\sum_{i=1}^{n} f(\xi_i)\Delta g_i
-
\sum_{i=1}^{n} f(\xi_i) g'(\xi_i)\Delta x_i
\right|
\le
M \sum_{i=1}^{n} | g'(t_i) - g'(\xi_i) | \Delta x_i
< M\varepsilon.
\]

Hence
\[
\left| \int_a^b f \, dg - \int_a^b f(x)g'(x)\,dx \right| \le M\varepsilon.
\]

Since $$\varepsilon$$ is arbitrary,
\[
\int_a^b f \, dg = \int_a^b f(x)g'(x)\,dx.
\]

**Second part.**
The mass distribution on $$[a,x]$$ is
\[
m(x) = \int_a^x \rho(t)\,dt.
\]

By the Fundamental Theorem of Calculus,
\[
m'(x) = \rho(x).
\]

Letting $$m(x) = g(x)$$,
\[
\int_a^b f \, dm = \int_a^b f(x)m'(x)\,dx,
\]
and hence the total mass is
\[
M = \int_a^b \rho(x)\,dx.
\]

(Here $$f(x)$$ is the constant function $$1$$.)

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Problem #9