If $\int_a ^b f df=0$ and f is continuous, then f is the function constant $0$ I have been studied some properties of Riemann Stieltjes integral, and i found this: If  $\int_a ^b f df=0$ and f is continuous for every $a<b$ in $R$, then f is the function constant $0$ without proof
I know, by definition $\int_a ^b f df = \sum_{i=0} ^{n} f(\lambda_i)(f(x_{i})-f(x_{i-1}))$ 
and if $f$ is constant, is clear  $(f(x_{i})-f(x_{i-1}))=0$ for every $i$
but why $f$ must be a constant zero?
 A: Nothing about $f$ being monotone or of bounded variation is specified here.  These are the typical assumptions for Riemann-Stieltjes integrators.  That limits the available tools for proving this.
Nevertheless, stating that $\int_a^b f \, df = 0$ for all $a < b$ implies that $f$ can be assumed to be RS integrable with respect to $f$.
For any $\epsilon > 0$, there exists a partition $P: a = x_0 < x_1 < \ldots < x_n = b$ such that for any choice of intermediate points $\xi_j \in [x_{j-1},x_j]$,
$$\left|\sum_{j=1}^n f(\xi_j)(f(x_j) - f(x_{j-1})) \right| < \frac{\epsilon}{2}$$
Thus,
$$\left|\sum_{j=1}^n (f(x_j)+f(x_{j-1}))(f(x_j) - f(x_{j-1})) \right|\leqslant \left|\sum_{j=1}^n f(x_{j-1})(f(x_j) - f(x_{j-1}) \right|+\left|\sum_{j=1}^n f(x_j)(f(x_j) - f(x_{j-1})) \right| < \epsilon,$$
and, 
$$\left|\sum_{j=1}^n (f(x_j)+f(x_{j-1}))(f(x_j) - f(x_{j-1})) \right| = \left|\sum_{j=1}^n ((f(x_j))^2 - (f(x_{j-1}))^2)\right| = \left|(f(b))^2 - (f(a))^2\right| < \epsilon$$
This implies that $(f(b))^2 = (f(a))^2$ and $f(b) = \pm f(a)$.
Since, $f$ is continuous, taking $a$ close to $b$ rules out $f(b) = -f(a)$.
The argument can be extended, proving that $f(a)$ is constant for all $a \in \mathbb{R}.$
Addendum
Also, the question of whether or not $0$ is the only admissible constant is worth considering.  Assume that for some fixed point $a$, we have $f(a) > 0$. Then for any other distant point $b$ we immediately know that $f(b) = \pm f(a)$. If $f(b) < 0$ then by continuity and the intermediate value property there would have to exist a point $c$ between $a$ and $b$ where $f(c) = 0$. But $f(c) = \pm f(a)$ and this would give us the contradiction that $f(a) = 0$. Thus, $f(x) = f(a)$ for all $x \in \mathbb{R}$.
A: (This is essentially RRL's argument, only simplified using “integration by parts,” which is a property of Riemann-Stieltjes integrals.)
If $f$ is Riemann-Stieltjes integrable with respect to itself on $[a, b]$ then integration by parts gives
$$
 \int_a^b f \, df = f^2(b) - f^2(a) - \int_a^b f \, df \\
\implies f^2(b) - f^2(a) = \frac 12 \int_a^b f \, df \, .
$$
It follows that $\int_a^b f \, df = 0$ for all $a < b$ if and only if $f^2$ is constant on $\Bbb R$. If $f$ is continuous then this is equivalent to $f$ being constant on $\Bbb R$.
As you already observed, the constant value is not necessarily zero.
