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

*

*The support supp$f_\theta (\mathbf{y})$ does not depend on $\theta$;

*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.

This is the definition of unbiasedness.
