Alright, let's tackle this Riemann-Stieltjes integration problem thoroughly, building on the fact you already established: $\gamma(x) = \int_a^x g , d\beta$ is continuous and of bounded variation on $[a,b]$.

1. Prove $\int_a^b f , d\gamma$ exists #

Since $f$ is continuous on the closed interval $[a,b]$, and we know $\gamma$ is a function of bounded variation on $[a,b]$, we can directly apply a key theorem from Rudin's Principles of Mathematical Analysis:

A continuous function on a closed interval is Riemann-Stieltjes integrable with respect to any function of bounded variation on that interval.

To break this down intuitively: bounded variation functions don't oscillate wildly, and continuous functions are uniformly continuous on compact intervals—these two properties together ensure that the Riemann-Stieltjes sums will converge to a unique limit as the mesh of the partition goes to 0. So $\int_a^b f , d\gamma$ is well-defined.

2. Prove $\int_a^b f , d\gamma = \int_a^b fg , d\beta$ #

We'll use the definition of Riemann-Stieltjes integrals and properties of the integral $\gamma(x)$ to connect the two integrals.

Step 1: Start with the Riemann sum for $\int_a^b f , d\gamma$ #

Take any partition $P = {a = x_0 < x_1 < \dots < x_n = b}$ of $[a,b]$, and choose sample points $t_i \in [x_{i-1}, x_i]$ for each $i$. The Riemann sum for $\int_a^b f , d\gamma$ is:
$$
S(P, f, \gamma) = \sum_{i=1}^n f(t_i) \left( \gamma(x_i) - \gamma(x_{i-1}) \right)
$$

Step 2: Substitute the definition of $\gamma$ #

By the definition of $\gamma$, $\gamma(x_i) - \gamma(x_{i-1}) = \int_{x_{i-1}}^{x_i} g , d\beta$. Substitute this into the sum:
$$
S(P, f, \gamma) = \sum_{i=1}^n f(t_i) \int_{x_{i-1}}^{x_i} g , d\beta = \sum_{i=1}^n \int_{x_{i-1}}^{x_i} f(t_i)g , d\beta
$$

Step 3: Show the sum converges to $\int_a^b fg , d\beta$ #

Since $f$ is uniformly continuous on $[a,b]$, for any $\epsilon > 0$, there exists a $\delta > 0$ such that if the mesh of $P$ is less than $\delta$, then $|f(x) - f(t_i)| < \frac{\epsilon}{M + 1}$ for all $x \in [x_{i-1}, x_i]$ (where $M = \int_a^b |g| , d\beta$, a finite value because $g \in \mathcal{R}(\beta)$).

Compute the difference between our sum and the target integral:
$$
\left| S(P, f, \gamma) - \int_a^b fg , d\beta \right| = \left| \sum_{i=1}^n \int_{x_{i-1}}^{x_i} \left( f(t_i) - f(x) \right)g(x) , d\beta(x) \right|
$$
Using the triangle inequality for integrals:
$$
\leq \sum_{i=1}^n \int_{x_{i-1}}^{x_i} |f(t_i) - f(x)| |g(x)| , d\beta(x)
$$
Substitute our uniform continuity bound:
$$
\leq \frac{\epsilon}{M + 1} \sum_{i=1}^n \int_{x_{i-1}}^{x_i} |g(x)| , d\beta(x) = \frac{\epsilon}{M + 1} M < \epsilon
$$

Step 4: Conclude equality #

As the mesh of $P$ approaches 0, $S(P, f, \gamma)$ converges to $\int_a^b f , d\gamma$, and the difference between this sum and $\int_a^b fg , d\beta$ goes to 0. Therefore:
$$
\int_a^b f , d\gamma = \int_a^b fg , d\beta
$$

Alternatively, you could use Rudin's integration by parts theorem for Riemann-Stieltjes integrals: since $\gamma$ is continuous and $f$ is continuous (hence integrable with respect to $\gamma$), we have:
$$
\int_a^b f , d\gamma = f(b)\gamma(b) - f(a)\gamma(a) - \int_a^b \gamma , df
$$
Since $\gamma(a) = 0$, this simplifies to $f(b)\int_a^b g , d\beta - \int_a^b \gamma , df$. Evaluating the remaining integral would lead you to the same equality, though the Riemann sum approach is more direct here.

内容的提问来源于stack exchange，提问作者user534253