Let $f$ continuous on $(-\infty,\infty)$, $\phi$ of bounded variation on $(-\infty,\infty)$, and $\lim_{|x|\to\infty}\ f(x)=0$ then the following Riemann-Stieltjes integral exists: $$\int_{-\infty}^{+\infty}f \ d\phi$$
A function $\phi$ is of bounded variation on $[a,b]$ if $\Gamma$ partition of $[a,b]$ and $\sup_\Gamma \sum_{i=0}^m |\phi(x_i)-\phi(x_{i-1})|=V(\phi;a,b)<\infty$ or equivalently if there is two increasing functions such that $\phi=\phi_1-\phi_2$. The $V(\phi, -\infty,\infty)=\lim_{a\to-\infty,b\to\infty}V(\phi;a,b)$.
Exercise 2.17 Measure and integral Intro to Real Analysis Wheeden & Zygmund
My proof: One may assume that $\phi$ is non decreasing. Let $\varepsilon>0$. For any $M>0$, the integral $\int_{-M}^{M}f \ d\phi$ exists, so it's enough to show there is $N>0$ such that for any $x>N$, the integral $\left|\int_N^x f d\phi\right|<\varepsilon$ and for any $x<-N$, the integral $\left|\int_{x}^{-N} f d\phi\right|<\varepsilon$.
Let's see the first: There is $M>0$ such that if $y>M$ then $|f(y)|<\frac{\varepsilon}{V(\phi;-\infty,\infty)}$ assumming the variation of $\phi$ is not $0$. Then picking $N=M+1$ by the mean-value theorem there is $\xi\in[M+1,x]$ such that $$0\leq \left|\int_{M+1}^x f d\phi\right|=\left|f(\xi)(\phi(x)-\phi(M+1)) \right|\leq \left|f(\xi)\right| \left|(\phi(x)-\phi(M+1)) \right|<\varepsilon $$ Therefore $\int_{M+1}^\infty f d\phi=0$. Similarly one can see $\int_{-\infty}^{-M-1} f d\phi=0$. So it exists $$\int_{-\infty}^\infty f\ d\phi=\int_{-M-1}^{M+1} f\ d\phi$$
I think this proof works (if not please let me known), the sandwich did the work.
Questions starts here
I have 2 questions:
(1) The reason for $\lim_{|x|\to \infty}f(x)=0$ to be needed, and couldn't be relaxed to $$\lim_{|x|\to \infty}f(x)=a$$ is because we get, for $h=\sup_{\mathbb{R}}{|f(x)|}$ $$0\leq \left|\int_{M+1}^x f d\phi\right|\leq h V(\phi,-\infty,\infty)$$ so no sandwich can be apply. It should leads to some argument to prove its divergence.
(2) Is there a proof without using the mean value theorem, using for example upper and lower sums, to prove that $0\leq U_\Gamma-L_\Gamma<\varepsilon $ or for example using Cauchy condition $|R_\Gamma-R_{\Gamma'}|<\varepsilon$ for any $|\Gamma,\Gamma'|<\delta$. If so, how to approach in this infinite improper case?
In this context $$R_\Gamma = \sum_{i=1}^m f(\xi_i)(\phi(x_i)-\phi(x_{i-1})),\ \ $$ $$ U_\Gamma=\sum_{i=1}^m\left(\sup_{[x_{i-1},x_i]}f \right)(\phi(x_i)-\phi(x_{i-1}))\ \mathrm{and}\ L_\Gamma = \sum_{i=1}^m\left(\inf_{[x_{i-1},x_i]}f \right)(\phi(x_i)-\phi(x_{i-1}))$$ so assuming $\phi$ increasing, $$L_\Gamma\leq R_\Gamma\leq U_\Gamma$$