# Show that the integrator in a Lebesgue-Stieltjes integration is zero

Let's consider a continuous function of bounded variation that is defined on a finite interval: $f: [0,t] \rightarrow \mathbb{R}$. Assume that we have a positive, continuous function that is bounded away from zero: $g: [0,t] \rightarrow \mathbb{R}$. Then we have the Lebesgue-Stieltjes integral of g with respect to f, well defined as a function of the upper limit of integration: $$I(s)=\int_{0}^{s}g(u)df(u), \ 0\leq s \leq t.$$ If we assume that $f(0)=0$ and that $I(s)=0, \forall 0 \leq s \leq t$, can we conclude that $f(s)=0, \forall s \in [0,t]$ ? Any help or comment is greatly appreciated, thanks!

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Don't you have the assumption that $f$ is left-continuous? –  Tunococ Aug 24 '12 at 6:08
@Tunococ I do have that $f$ is continuous. –  user7762 Aug 24 '12 at 14:13

The Lebesgue-Stieltjes integral $\int_0^x g(s)\ df(s) = \int_0^x g(s) d\mu(s)$ where $\mu$ is the signed measure on $[0,t]$ corresponding to $f$. By the Hahn decomposition theorem we can write $\mu = \mu_+ - \mu_-$ where $\mu_+$ and $\mu_-$ are finite positive measures, and there is a measurable $A \subseteq [0,t]$ such that $\mu_-(A) = 0$ and $\mu_+([0,t] \backslash A) = 0$. Thus $\mu_+(B) = \mu(A \cap B)$ and $\mu_-(B) = - \mu(B \backslash A)$ for all measurable $B$. Now if $f$ is not identically $0$, neither is $\mu$. If $V$ is the total variation of $f$ on $[0,t]$, $\|\mu\| = \|\mu_+\| + \|\mu_-\| = V > 0$.
Now let $g$ be a continuous strictly positive function on $[0,t]$. Thus
$B > g > \epsilon > 0$ for some constants $B$, $\epsilon$. We have $\int_0^t g \ d\mu_+ + \int_0^t g \ d\mu_- > \epsilon (\|\mu_+\| + \|\mu_-\|) = \epsilon V$, so $\max(\int_0^t g \ d\mu_+, \int_0^t g \ d\mu_-) > \epsilon V/2$. For definiteness suppose $\int_0^t g \ d\mu_+ > \epsilon V/2$ (the case with $\int_0^t g \ d\mu_- > \epsilon V/2$ will be similar). By Borel regularity there is an open set $U$ with $A \subseteq U$ and $\mu_-(U) < \epsilon V/(4 B)$. Thus $$\int_U g \ df = \int_A g \ d\mu_+ - \int_{U \backslash A} g \ d\mu_- > \epsilon V/2 - \epsilon V/4 > 0$$ But $U$ is the union of at most countably many disjoint open intervals $(a_i, b_i)$ and $\int_U g \ df = \sum_i \int_{a_i}^{b_i} g \ df = \sum_i (I(b_i) - I(a_i))$, so at least one $I(b_i) - I(a_i)$ must be nonzero.