# Riemann-Stieltjes Integral is Everywhere Zero

Suppose that $f:[a, b] \to \mathbb{R}$ is a continuous function such that $\int_a^x f(t)dt = 0$ for all $a \leq x \leq b$. Show that $f(x) = 0$ for all $x \in [a, b]$.

I assumed by contradiction that $f(x') = c \neq 0$ for some $x' \in [a, b]$. Then since $f$ is integrable on $[a, x']$, there is some partition $P_1^*$ such that for any refinement $P_1 = \{a = x_0 < \cdots x_n = x'\}$ of $P_1^*$ and any choice of $T$ such that $t_i \in [x_{i - 1}, x_i]$ we have that $|S(f, P_1, T)| < \varepsilon$. Next, since $f$ is integrable on $[a, x_{n - 1}]$ we can find some partition $P_2^*$ such that for any refinement $P_2$ of $P_2^*$ and we have that $|S(f, P_2, T)| < \varepsilon$. Then if $P = P_1 \cup P_2 = \{a = y_0 < \cdots < y_m = x'\}$, $$|S(f, P, T)| = \left|\sum_{i = 1}^m f(t_i) \Delta t_i\right| = \left|f(t_m) \Delta t_m + \sum_{i = 1}^{m - 1} f(t_i) \Delta t_i\right| \geq \left||f(t_m) \Delta t_m| - \left|\sum_{i = 1}^{m - 1} f(t_i) \Delta t_i\right|\right|$$

I think that I am headed in the right direction but I can't seem to make the inequality work out. I was wondering if I could get a hint.

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Hint: Use the fact that $f$ is not just integrable, but continuous. In particular, if there is some $x$ such that $f(x) > 0$, there is also an interval $I$ with $x \in I$ such that $f(y) > 0$ for all $y \in I$ (similar for $f(x) < 0$). – Johannes Kloos Apr 5 '13 at 16:49
Any particular reason why you referred to this as the Riemann–Stieltjes integral when you're specialising to integrating against the identity (and thus you just have the trusty old Riemann integral) rather than an arbitrary function $\alpha \in BV[a,b]$? – kahen Apr 5 '13 at 16:53
I believe I'm supposed to prove this using the criterion for integrability for Riemann-Stieltjes integrals with $\alpha(t) = t$ as opposed to using lower and upper limits as in the case of a simple Riemann integral. – icaruss Apr 5 '13 at 16:58
Sorry guys, I'm still a bit confused as to how to set up the inequality. I know that I'll need to show that $|S(f, P, T)| \geq \varepsilon$, and that I can choose some partition with endpoints from the interval as pointed out by Johannes, but I'm not sure where to go from there. All I can think of is the reverse triangle inequality. – icaruss Apr 5 '13 at 17:03

Hint: Prove the contrapositive. Suppose $f\neq0$ then there is a smallest number $c$ in the support of $f$. You will have $\int_a^c f(x)\,dx = 0$ because $f(x)=0$ for $a\leq x < c$. You also have that $|f|$ is positive on an entire interval $(c,c+\delta)$. Now show that
$$\int_a^{c+\delta}f(x)\,dx = \int_a^c f(x)\,dx + \int_c^{c+\delta} f(x)\,dx \neq 0.$$
When $f$ is not identically $0$ on $[a,b]$ there is a $\xi\in[a,b]$ with $f(\xi)=:c\ne0$. Assume without restriction of generality that $c>0$. Since $f$ is continuous there is a $\delta>0$ and an interval $I:=[a',b']\subset[a,b]$ of length $\delta$ such that $a'\leq\xi\leq b'$ and $f(t)\geq {c\over2}$ for all $t\in I$. It follows that $$\int_a^{b'} f(t)\ dt-\int_a^{a'} f(t)\ dt=\int_{a'}^{b'} f(t)\ dt\geq {c\over2}\>\delta>0\ .$$ Therefore the function $F(x):=\int_a^x f(t)\ dt$ cannot be identically $0$ on $[a,b]$.