Why does $T$ being an unbiased estimator for $g(\theta)$ imply that $g(\theta) = ET = \int T(\mathbf{y}) f_\theta(\mathbf{y}) \ d\mathbf{y}$?

Question

I am currently studying the Cramer-Rao lower bound. My notes say the following:

Theorem: Cramer-Rao lower bound
Let $Y_1, \dots, Y_n$ have a joint distribution $f_\theta (\mathbf{y})$, where $f_\theta (\mathbf{y})$ satisfies the following two regularity conditions:

1. The support supp$f_\theta (\mathbf{y})$ does not depend on $\theta$;
2. for any statistic $T = T(Y_1, \dots, Y_n)$ with a finite $E_\theta \lvert T \rvert$, integration w.r.t. $\mathbf{y}$ and differentiation w.r.t. $\theta$ in calculating $(E[T])_\theta^\prime$ can be interchanged, that is, $$(E[T])_\theta^\prime = \dfrac{d}{d\theta} \int T(\mathbf{y})f_\theta (\mathbf{y}) \ d\mathbf{y} = \int T(\mathbf{y})f_\theta^\prime (\mathbf{y}) \ d\mathbf{y}$$ Let $T$ be an unbiased estimator for $\theta$, with a finite variance. Then $$\text{Var}(T) \ge \dfrac{1}{I(\theta)},$$ where $I(\theta) = E[(\ln(f_\theta(\mathbf{Y})_\theta^\prime)^2]$ is called the Fisher information number.
   More generally, if $T$ is an unbiased estimator for $g(\theta)$, where $g(\cdot)$ is differentiable, then, $$\text{Var}(T) \ge \dfrac{(g^\prime(\theta))^2}{I(\theta)}$$
   Proof: Cramer-Rao lower bound
   Since $T$ is an unbiased estimator for $g(\theta)$, we have $$g(\theta) = ET = \int T(\mathbf{y}) f_\theta(\mathbf{y}) \ d\mathbf{y},$$ and, under regularity conditions, $$g^\prime(\theta) = (E[T])_\theta^\prime = \dfrac{d}{d \theta} \int T(\mathbf{y}) f_\theta(\mathbf{y}) \ d\mathbf{y} = \int T(\mathbf{y}) f_\theta^\prime (\mathbf{y}) \ d\mathbf{y} = \int T(\mathbf{y})(\ln \left[ f_\theta (\mathbf{y})\right]_\theta^\prime f_\theta (\mathbf{y}) \ d\mathbf{y} = E[T(\mathbf{Y})(\ln [f_\theta(\mathbf{Y})])_\theta^\prime ]$$

I am confused by this part:

Since $T$ is an unbiased estimator for $g(\theta)$, we have $$g(\theta) = ET = \int T(\mathbf{y}) f_\theta(\mathbf{y}) \ d\mathbf{y}$$

Why does $T$ being an unbiased estimator for $g(\theta)$ imply that $g(\theta) = ET = \int T(\mathbf{y}) f_\theta(\mathbf{y}) \ d\mathbf{y}$?

---

EDIT

From https://en.wikipedia.org/wiki/Bias_of_an_estimator#Definition :

"... or equivalently, if the expected value of the estimator matches that of the parameter." In this case, the expected value of the estimator does not match the parameter, but rather a function of the parameter. So there seems to be some disconnect here between what's in these notes and the Wikipedia definition of an unbiased estimator.

Answer 1

This is the definition of unbiasedness.