I am trying to evaluate the plug-in estimator $$\hat{\theta} = T(F_n)$$ where $F_n (t) = \frac{1}{n} \sum_{i = 1}^{n}1\{t_i \leq t\}$ and $T : \eta \to \int_{}^{}x \ d\eta(x) $, (Riemann-Stieltjes integral).
My approach has been to try to use "Lemma 7.4" found on page 134 in "A Concise introduction to mathematical statistics" by Dragi Anevski which states that if $F$ is an increasing step function on $I$, so that $$F(t) = \sum_{i=1}^{N}a_i \ 1\{t\leq t_i\}$$ with $min(I) = t_0 < t_1 ... <t_N = max(I)$ and $a_i \geq 0$. Then if $g$ is continuous, $$\int_{I}^{}g(x)dF(x) = \sum_{i=1}^{N} g(t_i)a_i$$.
So if we let $g(x) = x$, $a_i = 1$ and $$F_n(x) = \frac{1}{n}\sum_{i=1}^{n}1\{x_i \leq x\} = \frac{1}{n}\sum_{i=1}^{n}1\{\frac{1}{x} \leq \frac{1}{x_i}\} = \frac{1}{n}\sum_{i=1}^{n}1\{t \leq t_i\}$$, with $t = \frac{1}{x}$ and $t_i = \frac{1}{x_i}$, then we have that $$\hat{\theta} = T(F_n) = \int_{}^{}x \ d{F_n}(x) = \frac{1}{n}\sum_{i=1}^{n}\frac{1}{x_i}$$.
This is apparently wrong because the very same book says it should be $\frac{1}{n}\sum_{i=1}^{n}x_i$. Why is this true and where do I go wrong? I found another post on here about this calculation but it involved the Dirac function and I have not yet learned about it.
Thanks in advance, $$\\$$ Isak
There seems to be an error either in that lemma or in your rendition of it. The function
$$ F(t) = \sum_{i=1}^{N}a_i \ 1\{t\le t_i\} $$
with $a_i\gt0$ is a decreasing step function (the value at $-\infty$ is $\sum_ia_i$ and the value at $+\infty$ is $0$). A correct formulation of the lemma would be for an increasing step function $F$ of the form
$$ F(t) = \sum_{i=1}^{N}a_i \ 1\{t\ge t_i\}\;. $$
Then you don't need to turn the inequality around, and the result comes out right.
Your manipulation involving $t$ and $x$ isn't valid because you applied the lemma for $t$ even though the measure is $\mathrm dF_n(x)$, not $\mathrm dF_n(t)$. A valid way to turn around the inequality (if there'd been a need for turning it around) would have been to use $1\{t\ge t_i\}=1-1\{t\lt t_i\}$ and thus $\mathrm d1\{t\ge t_i\}=-\mathrm d1\{t\lt t_i\}$.